The following information regarding a portfolio of two stocks are given: w1 = .25, w2 = .75, E(R1) = .08, and E(R2) = .15.

Which of the following regarding the portfolio expected return, E(Rp), is correct?
-.3640
-.2300
-.1325
-.1699

Answer:

At 0.05 significance level, the p-value is 0.014

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 3 minutes

Sample mean, [tex]\bar{x}[/tex] = 3.11 minutes

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, σ = 0.5 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 3\text{ minutes}\\H_A: \mu > 3\text{ minutes}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

[tex]z_{stat} = 2.2[/tex]

Now, we calculate the p-value from the normal standard z-table.

P-value = 0.014

At 0.05 significance level, the p-value is 0.014

We can use the inverse function derivative theorem:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.[/tex]

In this case, we want to evaluate [tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}[/tex], so:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.[/tex]

The derivative is:

[tex]\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.[/tex]

The ordinate of the point is [tex]f^{-1}(1) = 0[/tex], so we evaluate:

[tex]\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.[/tex]

Finally:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.[/tex]

We can check the answer by finding the inverse:

[tex]y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},[/tex]

so that

[tex]f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.[/tex]

Therefore:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.[/tex]

Which finally gives the same answer as before:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.[/tex]

Answer: [tex]\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.[/tex]

Final answer:

The equivalence class of 1 consists of all integers plus 1, and the equivalence class of 1/2 consists of all numbers of the form 1/2 plus any integer.

Explanation:

The equivalence relation R is defined such that (x, y) is in R if and only if x - y is an integer. For any real number a, the equivalence class of a is the set of all real numbers b such that a - b is an integer.

Equivalence Class of 1

The equivalence class of 1 includes all real numbers that are an integer distance from 1. This means it contains all numbers of the form 1 + k, where k is any integer. Hence, it includes numbers like 0, 1, 2, 3, and so on, in addition to negative integers: -1, -2, -3, etc.

Equivalence Class of 1/2

Similarly, the equivalence class of 1/2 consists of all real numbers of the form 1/2 + k, where k is any integer. This set includes numbers like -1/2, 1/2, 3/2, 5/2, and so on.

Answer:

Step-by-step explanation:

Linear relations are those which are represented by straight line on a graph i.e. increase or decrease a variable directly effect the other in a linear way

for example

y=2x+3

increase in value of x increase the value of y linearly.

Non-linear relationship are those in which change in one entity does not correspond to change in other quantity.

Real life situation for linear relation is when we apply force on a block it accelerates and it goes on increasing as force is increasing.

For Non-linear relation

increase in temperature vs time spend

Time spent is more to increase the slight amount of temperature.            

Linear and non-linear systems exist in everyday scenarios. A linear example is cost increasing by a fixed amount per item purchased, while a non-linear example might be a room's temperature: heating up quickly initially, then more slowly as it approaches the desired warmth.

In everyday life, both linear and non-linear systems abound. A straightforward example of a linear system is calculating the total cost of goods: if each item costs the same, your total cost represents a linear relationship with the number of items - the cost increases by a set amount per item.

On the other hand, a non-linear system might relate to how the temperature of a heated room changes over time: it may start increasing rapidly, then slowly once the room approaches the desired temperature. This represents a non-linear relationship as the rate of temperature increase isn't constant.

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Answer: confidence interval = b. ( 116.42 to 123.58)

Step-by-step explanation:

Given;

Number of samples n = 30

Standard deviation r = 10

Mean x = 120

Confidence interval of 95%

Z' = t(0.025) = 1.96

Confidence interval = x +/- Z'(r/√n)

= 120 +/- 1.96(10/√30)

= 120 +/- 3.58

= ( 116.42, 123.58)

The subject of the question is Mathematics, relevant to High School students. It involves simplifying and combining radicals, eliminating terms to simplify algebraic expressions, and understanding transcendental numbers and their properties.

Add and Subtract Radicals

When working with radicals, it is essential to simplify each radical by removing perfect square roots in order to combine like radicals effectively. For instance, to add \\(\\sqrt{18} + \\sqrt{8}\\), we must first simplify. \\(\\sqrt{18}\\) becomes \\(\\sqrt{9*2}\\) or \\(\\sqrt{9}\\cdot\\sqrt{2}\\), which simplifies to \\(\\sqrt{2}\\) times 3. Similarly, \\(\\sqrt{8}\\) can be rewritten as \\(\\sqrt{4*2}\\) or \\(\\sqrt{4}\\cdot\\sqrt{2}\\), which simplifies to \\(\\sqrt{2}\\) times 2. Now we have like radicals and can combine them: 3\\(\\sqrt{2}\\) + 2\\(\\sqrt{2}\\) equals 5\\(\\sqrt{2}\\).

Simplify Algebra and Reasonableness

To simplify the algebra, we identify and eliminate terms where possible, checking for reasonableness of the answer at all times. Verify if the operations you have performed are correct and the solution looks reasonable given the original equation or expression.

Transcendental Numbers and Functions

Transcendental numbers, like \\(\\sqrt{2}\\) or \\(\\sqrt{5}\\), are those that are not the root of any non-zero polynomial equation with rational coefficients. In algebra, these play a vital role in understanding real numbers and theorems related to them.

Answer:

Step-by-step explanation:

Given that the Products produced by a machine has a 3% defective rate.

Each product is independent of the other with a constant prob of being defective as 0.03

X - no of defects is binomial with p =0.03

a)  the probability that the first defective occurs in the fifth item inspected

=Prob for first 4 non defective and 5th defective

=[tex](0.97)^4(0.03)^1\\=0.0266[/tex]

(b) the probability that the first defective occurs in the first five inspections

=P(X=1) in binomial with n=5

= 0.9915

c) the expected number of inspections before the first defective occurs

Expected defects in n trials = np

Expected number of inspection before the first defect = 1/p

= 33.33

=34

(c) What is the expected number of inspections before the first defective occurs?

Final answer:

The probability that the first defective product occurs on the fifth item is approximately 0.0895. The probability that a defective product occurs within the first five inspections is approximately 0.1426. The expected number of inspections before a defective product occurs is around 33 items.

Explanation:

To find the probability that the first defective occurs on the fifth item inspected (part a), we consider the first four items to be non-defective and the fifth one to be defective. The probability for one non-defective item is 97%, or 0.97. The probability we are looking for is thus the product of these probabilities:

P(non-defective, non-defective, non-defective, non-defective, defective) = (0.97)⁴* 0.03 ≈ 0.0895

For part b, we want to find the probability that a defective is found at any point in the first five inspections. This is the sum of the probabilities of finding the first defective on the first, second, third, fourth, or fifth inspection:

P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th) = 0.03 + (0.97 * 0.03) + (0.97² * 0.03) + (0.97³ * 0.03) + (0.97⁴ * 0.03) ≈ 0.1426

For part c, the expected number of inspections before finding the first defective item is given by the formula E(X) = 1/p, where p is the probability of finding a defective item. Substituting the given probability:

E(X) = 1/0.03 ≈ 33.33

Therefore, we expect to inspect approximately 33 items before finding the first defective.

Answer:

b. 85

Step-by-step explanation:

Average grade (μ) = 75

Standard deviation (σ) = 8

Assuming a normal distribution, the z-score corresponding to the upper 10% of the test grades is z = 1.28.

The minimum grade 'X' within the top 10% is given by:

[tex]z=\frac{X-\mu}{\sigma}\\1.28=\frac{X-75}{8}\\X=85.24[/tex]

Rounding to the nearest percent, the dividing point between an A and a B grade is 85.

Final answer:

The dividing point between an A and B grade for the upper 10% of the test grades is calculated using the normal distribution properties. A Z-score of 1.28 corresponds to the 90th percentile, resulting in a score of 85.24, which is rounded to 85 according to standard rounding rules.

Explanation:

To determine the dividing point between an A and B grade for the upper 10% of the test grades in History 101, we need to refer to the properties of the normal distribution. The mean score is given as 75 with a standard deviation of 8. We are looking for the score that corresponds to the 90th percentile since the instructor wants to award an A to the upper 10%. Z-scores allow us to translate percentile ranks into scores on a given normal distribution.

Using a standard normal distribution table, we find that a Z-score of approximately 1.28 corresponds to the 90th percentile. To find the actual score, we use the formula Score = Mean + (Z-score imes Standard Deviation). Plugging in the values:

Mean = 75

Standard Deviation = 8

Z-score for the 90th percentile = 1.28

Score = 75 + (1.28 imes 8) = 75 + 10.24 = 85.24

When applying the standard rules for rounding, we round 85.24 to the nearest whole number, which is 85. Therefore, the dividing point between an A and a B grade is an 85. Students need to score at this point or higher to be in the top 10% and receive an A grade.

Answer:

[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]      

[tex]p_v =P(t_{15}>2)=0.0320[/tex]  

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.  

Step-by-step explanation:

1) Data given and notation      

[tex]\bar X=25[/tex] represent the sample mean      

[tex]s=2[/tex] represent the standard deviation for the sample      

[tex]n=16[/tex] sample size      

[tex]\mu_o =24[/tex] represent the value that we want to test    

[tex]\alpha[/tex] represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

Confidence =0.95 or 95%

[tex]\alpha=0.05[/tex]

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \leq 24[/tex]      

Alternative hypothesis:[tex]\mu > 24[/tex]      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=16-1=15[/tex]  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{15}>2)=0.0320[/tex]  

Conclusion      

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.      

Answer: x = 120 degrees

Step-by-step explanation:

The diagram is that of a polygon with 5 sides. This means that it is a Pentagon. The sum of the interior angles in a polygon is expressed as

180(n -2)

Where n represents the number of sides that the polygon has.

Since the given polygon has 5 sides, then the sum of the interior angles would be

180(5 - 2) = 180 × 3 = 540 degrees.

Therefore,

x + x + x + 90 + 90 = 540

3x + 180 = 540

3x = 540 - 180 = 360

x = 360/3 = 120 degrees

Answer:

x = 4 + (log 6 / log 3)

x ≈ 5.631

Step-by-step explanation:

3^(x − 4) = 6

Take log base 3 of both sides.

log₃ 3^(x − 4) = log₃ 6

x − 4 = log₃ 6

Use change of base formula.

x − 4 = log 6 / log 3

Solve for x.

x = 4 + (log 6 / log 3)

x ≈ 5.631

Answer:

5.631

Step-by-step explanation:

Using the change of base formula log base b of y equals log y over log b

Log y (base b) = log y /log b

3^(x − 4) = 6

Taking the log of both sides

log 3^(x − 4) = log 6

using the logarithm law that states that

log a ^ x = x log a

x - 4 log 3 = log 6

x - 4 = log 6 / log 3

x - 4 = 1.630929754

x = 5.630929754

≈ 5.631

Answer:

Matching the vocabulary word with its corresponding example:

1. All children who Ski and Snowboard = population (a group of items, units or subjects which is under reference of study e.g inhabitants of a region, numbers of cars in a city e.t.c)

2. The 92 children who were asked when they took their first lesson = Sample (a part or fraction of a population selected on some basis)

3. The average age that all children take their first to lesson = Parameter (the number that summarizes some characteristics of a population

4. The average age that the 92 children took their first lesson = statistic (a sample characteristic corresponding to the population parameter used when a sample is use to make inference about a population)

5. The age that children take their first listen = variable (anything that has attribute, quality or quantity that varies or a characteristic/attribute that describes a place, person or thing)  

6. The list of the 92 ages that the children from the study took the first listen = data (facts that are collected together for analysis)

Step-by-step explanation:

1. All children who Ski and Snowboard = population (a group of items, units or subjects which is under reference of study , numbers of cars in a city e.t.c)

2. The 92 children who were asked when they took their first lesson = Sample (a part of a population selected on some basis)

3. The average age that all children take their first to lesson = Parameter (the number that summarizes some characteristics of a population

4. The average age that the 92 children took their first lesson = statistic (a sample characteristic corresponding to the population parameter used when a sample is use to make inference about a population)

5. The age that children take their first listen = variable (anything that has attribute, quality or quantity that varies or a characteristic/attribute that describes a place, person or thing)  

6. The list of the 92 ages = data (facts that are collected together for analysis)

Answer:

False. The products from cross multiplication are different.

Step-by-step explanation:

To know if a pair of ratios form a proportion, cross multiply. If the products are equal, they are a proportion.

Write like this to see top (numerator) and bottom (denominator) clearly.

[tex]\frac{4}{5} =\frac{5}{6}[/tex]

Multiply each numerator with the other side's denominator:

4 X 6 = 24

5 X 5 = 25

Are they equal? No. 24 ≠ 25

Therefore it's not a proportion.

Answer:

(a) 0.12

(b) 0.42

(c) 0.46

Step-by-step explanation:

Probability of student A not showing up = 0.3

Probability of student B not showing up = 0.4

(a) Neither will show up for class

[tex]P(A\ and\ B) = 0.3*0.4 = 0.12[/tex]

(b) Both will show up for class

[tex]P (A\ nor\ B) = (1-0.3)*(1-0.4) = 0.42[/tex]

(c) Exactly one will show up for class

[tex]P(A\ or\ B) = 1 -P(A\ and\ B) - P(A\ nor\ B)\\P(A\ or\ B) = 1 -0.12-0.42 = 0.46[/tex]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The correct statement is that the middle 50% of personal injury claim costs fall between $989.72 and $1,088.18, which represents the interquartile range. This range and the standard deviation are key in evaluating the distribution of claim costs.

The statement that the middle 50% of the costs are between $989.72 and $1,088.18 is correct in reference to the provided summary statistics of personal injury claims. This range is defined by the first and third quartiles, also known as the interquartile range (IQR). The IQR is a measure of variability and represents the span between the 25th percentile (first quartile) and the 75th percentile (third quartile), which indeed encompasses the middle 50% of data in a given sample.

In the context of personal injury claims costs at your firm, this means that half of the claim costs fall within that range, with fewer costs being less than $989.72 (the lower 25%) and fewer costs being more than $1,088.18 (the upper 25%). This can be useful information for assessing claims costs and preparing for future claims expenses. The provided standard deviation of $89.50 indicates the average amount that claim costs vary from the mean ($1,040.47).

Answer:

a) The 90% confidence interval would be given by (721.716;790.724)  

b) We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)  

c) We assume the following conditions:

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

[tex]\bar X=756.22[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]s=100.89[/tex] represent the sample standard deviation  

n=25 represent the sample size  

90% confidence interval  

The confidence interval for the mean is given by the following formula:  

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)  

The degrees of freedom are given by:

[tex]df=n-1=25-1=24[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,24)".And we see that [tex]t_{\alpha/2}=1.71[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]756.22-1.71\frac{100.89}{\sqrt{25}}=721.716[/tex]  

[tex]756.22+1.71\frac{100.89}{\sqrt{25}}=790.724[/tex]  

So on this case the 90% confidence interval would be given by (721.716;790.724)  

Part b

We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)  

Part c

We assume the following conditions:

Final answer:

The 90% confidence interval for the true speed of light is between 299,723.02683 km/sec and 299,789.41317 km/sec. This interval suggests we can be 90% confident that the constant speed of light falls within this range, with the understanding that the true speed of light is approximately 299,792,458 m/s.

Explanation:

The sample mean is 756.22, and the standard deviation is 100.89 with 25 trials.

First, add 299,000 km/sec to the sample mean to revert to the actual speed of light. Adjusted mean = 756.22 + 299,000 = 299,756.22 km/sec.

Since the sample size (25) is greater than 30, we use the z-score for a 90% confidence interval, which is 1.645.

To find the margin of error (ME), use the formula ME = z * (σ/√n), where σ is the standard deviation and n is the sample size. ME = 1.645 * (100.89/√25)= 1.645 * 20.178 = 33.19317 km/sec.

The confidence interval is then mean ± ME. That gives us the interval: [299,756.22 - 33.19317, 299,756.22 + 33.19317] or [299,723.02683, 299,789.41317] km/sec.

Interpretation: We are 90% confident that the true speed of light lies within the interval of 299,723.02683 to 299,789.41317 km/sec.

The samples are independent and randomly selected.

The data reported is accurate and measured without systematic errors.

The data is normally distributed or the sample size is large enough for the Central Limit Theorem to apply.

These results align with the known fact that the speed of light is a constant at approximately 299,792,458 meters/second, and any deviations observed in the experiment are likely due to measurement error or experimental uncertainties.

Answer:

Option A) About ±2.98 minutes

Step-by-step explanation:

We are given the following information in the question:

Sample mean, [tex]\bar{x}[/tex] = 191.3 minutes

Sample size, n = 200

Population standard deviation, σ = 21.5 minutes

Alpha, α = 0.05

The leaders of the organization wish to develop an interval estimate with 95 percent confidence.

[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]

Margin of error =

[tex]z_{\text{critical}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get:

[tex]\pm 1.96\times \displaystyle\frac{21.5}{\sqrt{200}} = \pm 2.9797 \approx \pm 2.98[/tex]

Option A) About ±2.98 minutes

Answer:

a) [tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]

b) [tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]

c) [tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]

Step-by-step explanation:

Let X the random variable that represent the number of airline crashes in a country. We know that [tex]X \sim Poisson(\lambda=2)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

And f(x)=0 for other case.

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

(a) What is the probability that there will be at least 3 accidents in the next month?

On this case we are interested on the probability of having at least three accidents in the next month, and using the complement rule we have this:

[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-2} 2^0}{0!}=0.1353[/tex]

[tex]P(X=1)=\frac{e^{-2} 2^1}{1!}=0.2707[/tex]

[tex]P(X=2)=\frac{e^{-2} 2^2}{2!}=0.2707[/tex]

And replacing we have this:

[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]

(b) What is the probability that there will be at least 6 accidents in the next two months?

For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x2= 4[/tex] accidents per 2 months.

[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-4} 4^0}{0!}=0.0183[/tex]

[tex]P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733[/tex]

[tex]P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465[/tex]

[tex]P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954[/tex]

[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]

[tex]P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563[/tex]

Replacing we got:

[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]

(c) What is the probability that there will be at most 6 accidents in the next three months?

For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x3= 6[/tex] accidents per 3 months.

[tex]P(X\leq 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)[/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-6} 6^0}{0!}=0.00248[/tex]

[tex]P(X=1)=\frac{e^{-6} 6^1}{1!}=0.0149[/tex]

[tex]P(X=2)=\frac{e^{-6} 6^2}{2!}=0.0446[/tex]

[tex]P(X=3)=\frac{e^{-6} 6^3}{3!}=0.0892[/tex]

[tex]P(X=4)=\frac{e^{-6} 6^4}{4!}=0.1339[/tex]

[tex]P(X=5)=\frac{e^{-6} 6^5}{5!}=0.1606[/tex]

[tex]P(X=6)=\frac{e^{-6} 6^6}{6!}=0.1606[/tex]

[tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]

Answer:

[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]

And that would be the approximately cost for 2013.

Step-by-step explanation:

For this case we need to define some notation first.

A= population , t= represent the years after 2003, C= constant for the exponential model.

The starting point t=0 correspond to the year of 2003.

On this case we are assuming the following exponential model:

[tex]A(t) = A_o e^{Ct}[/tex]

The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:

[tex]120=A_o e^{C(0)}=A_o e^0 = A_o[/tex]

And then the model is:

[tex]A(t) =120 e^{Ct}[/tex]

Now we need to determine the value for C. Since we know that inflation increase 4% per year we have that after one year we have 1.04 times the value of the original value, and we have this equation:

[tex]1.04 A_o= A_o e^{C(1)}= A_o e^C[/tex]

And we got this:

[tex]1.04= e^C [/tex]

Applying ln on both sides we got:

[tex]ln(1.04)= C=0.0392207[/tex]

So then our model is given by:

[tex]A(t) = 120 e^{ln(1.04)t}[/tex]

For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:

[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]

And that would be the approximately cost for 2013.

The cost of groceries in 2013, after applying an annual inflation rate of 4% for 10 years, will be approximately $177.63.

To calculate the annual rate of inflation and the cost in 2013, you can use the exponential growth function, where cost =[tex]initial_{cost} * (1 + rate)^{time[/tex]. In this case, the initial cost in 2003 (t=0) is $120 and the annual rate of inflation is 4%, or 0.04. To calculate the cost in 2013, which is 10 years after 2003, apply the exponential growth function:

Cost in 2013 = $120 * (1 + 0.04)¹⁰

This calculation yields:

Cost in 2013 = $120 * (1.04)¹⁰

Cost in 2013 = $120 * 1.48024

Cost in 2013 = $177.63

Therefore, in 2013, assuming the rate of inflation remains constant, the cart of groceries will cost approximately $177.63.

Answer:

The area of this revolted surface is 36π

Step-by-step explanation:

To obtain the area of a revolted surface, you have to define:

1) which is the axis on which the surface is revolted: this defines the limits on that axis or hight of the surface. In this case x∈[0;2]

2) which is the expression of the radius of the revolted surface and its dependence with the hight. In this case, the radius expression could be Y=4x+5

3) Define the angular variable: If this is a fully revolted surface, the angular variable will go from 0 to 2π

Now we can obtain the area with a double integral:

[tex]A=\int\limits^{2}_0 { \int\limits^{2\pi}_0 {r} \, d \varphi } \, dx =\int\limits^{2}_0 { \int\limits^{2\pi}_0 {4x+5} \, d \varphi } \, dx =\int\limits^{2}_0 { (2\pi)(4x+5)} \, dx=36\pi[/tex]

Answer: 17-7i

Step-by-step explanation:

Answer:

[tex]\oint_cF.dr=0\\[/tex]

Step-by-step explanation:

Given that a circle C of radius 7

[tex]x^{2} +y^{2} =49---(1)[/tex]

To find:

[tex]\oint_{C}F.dr[/tex]

As NO function is given so we suppose it to be:

[tex]F=<x,y>[/tex]

Parametric equations:

[tex]x=rcos\theta=7cos\theta\\y=rsin\theta=7sin\theta[/tex]

Each point on circle can be then found as

[tex]r(\theta)=<7cos\theta,7sin\theta>---(2)[/tex]

From (2) dr can be found as:

[tex]dr=<-7sin(\theta),7cos(\theta)>d\theta---(3)[/tex]

From (2) and (3)

[tex]\oint_cF.dr=\int_{0}^{2\pi}{<7cos\theta,7sin\theta><-7sin(\theta),7cos(\theta)>}\,d\theta\\\\\oint_cF.dr=\int_{0}^{2\pi}{<(-7cos\theta)(7sin\theta),(7sin(\theta))(7cos(\theta))>}\,d\theta\\\\\\\oint_cF.dr=\int_{0}^{2\pi}{-49cos\theta sin\theta+49sin(\theta)cos(\theta)}\,d\theta\\\\\oint_cF.dr=0\\[/tex]

Answer:

8x + 6y >/= 48 ......1

x + 2y >/= 12 .......2

The cost function is given as;

C = 0.05x + 0.03y .........3

The minimum cost is $0.24 at (0,8)

That is 0 ounces of oats and 8 ounces of rice.

Step-by-step explanation:

let x represent the number of ounces of oats

And y represent the number of ounces of rice

For Vitamin A

Minimum requirements = 48mg

x ounces of oats contribute 8mg × x

y ounces of Rice contribute 6mg × y

Therefore, we have;

8x + 6y >/= 48 ......1

For Vitamin B

Minimum requirements = 12mg

x ounces of oats contribute 1mg × x

y ounces of Rice contribute 2mg × y

Therefore, we have;

x + 2y >/= 12 .......2

The cost function is given as;

C = 0.05x + 0.03y .........3

Attached is the graphical representation.

The feasible points are (x,y) = (0,8),(2.4,4.8),(12,0)

The minimum cost is determined by substituting each point into the cost function

For (0,8)

C= 0.05(0) + 0.03(8)

C = $0.24

For (12,0)

C= 0.60

For (2.4,4.8)

C= $0.264

The minimum cost is $0.24 at (0,8)

The problem can be modelled with a system of linear inequalities to represent the constraints of the cereal company. You graph these constraints and find the feasible region. After graphing the objective function, move this line towards the origin until it just leaves the feasible region. This point gives the optimal solution.

In this problem, we are dealing with linear equations and inequalities. The goal of the Munchies Cereal Company is to determine the amount of oats and rice, measured in ounces, to include in its cereal mix so as to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while at the same time minimizing the cost.

Let's denote the amount of oats as 'x' and the amount of rice as 'y'. The nutrition constraints can be formulated as:

And since quantities cannot be negative, we also have the constraints: x >= 0 and y >= 0. The objective is to minimize the cost, which can be expressed as C = 0.05x + 0.03y.

To solve this problem graphically, you would plot the constraint lines and see the feasible region (the area that satisfies all constraints). The cost line (C = 0.05x + 0.03y) is then drawn and moved towards the origin until the last point of the feasible region is touched. That point gives the optimal solution.

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Answer:

(a) The fraction of employees is 0.84.

(b)

[tex]\mu=850\\\\\sigma=450[/tex]

(c)

[tex]\mu=787.5\\\\\sigma=472.5[/tex]

(d) No. The left part of the distribution would be truncated too much.

Step-by-step explanation:

(a) If the weekly salaries are normally​ distributed, estimate the fraction of employees that make more than ​$300 per week.

We have to calculate the z-value and compute the probability

[tex]z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84[/tex]

(b) If every employee receives a​ year-end bonus that adds ​$100 to the paycheck in the final​ week, how does this change the normal model for that​ week?

The mean of the salaries grows $100.

[tex]\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850[/tex]

The standard deviation stays the same ($450)

[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2}  } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2}  }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2}  } =\sigma[/tex]

(c) If every employee receives a 5​% salary increase for the next​ year, how does the normal model​ change?

The increases means a salary X is multiplied by 1.05 (1.05X)

The mean of the salaries grows 5%, to $787.5.

[tex]\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5[/tex]

The standard deviation increases by a 5% ($472.5)

[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2}  } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2}  }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5[/tex]

(d) If the lowest salary is ​$300 and the median salary is ​$525​, does a normal model appear​ appropriate?

No. The left part of the distribution would be truncated too much.

Normal distribution has its mean, median and mode coincident on single point. The solutions to the given problems are specified as:

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

For the given case, if we take X = salary of employees weekly of the considered company, then:

[tex]X \sim N(750, 450)[/tex]

The  fraction of employees that make more than ​$300 per week is

P(X > 300).

Using the standard normal distribution, we can estimate this  fraction of employees that make more than ​$300 per week as:

[tex]P(X > 300 ) = 1 - P(X \leq 300) = 1 - P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{300 - 750}{450} =-1)\\ \\P(X > 300) = 1 - P(Z \leq -1)\\[/tex]

Using the z-table, the p-value for Z = -1 is: 0.1587

Thus, [tex]P(X > 300) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413[/tex]

When we add $100 to each value of X, it doesn't change the structure of the graph of normal distribution.
When we increase salary by 5%, it means, new salary [tex]Y = X + 5\% \text{of X} = \dfrac{21X}{20}[/tex]

The random variable is scaled by 21/20 = 1.05, so the graph will stay on same origin but it will be stretched a bit in thickness and height.

For the 4th case(d), the median is specified to be $525, but as it was known to us that mean is $750, so mean and median aren't coinciding, and specially, the median is in left of mean, showing that the graph is leaning on right side(negatively skewed). (we deduce it when median < mean, as median shows that mid value is reached, but mean shows that probability is still not reached, so its being late, and reaches later, showing that there is tail in the left of the graph, so being negatively skewed).

Thus, The solutions to the given problems are specified as:

Learn more about standard normal distribution here:

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Answer: Addie weighs 4 ounces

Missy weighs 14 ounces

Corky weighs 8 ounces

Step-by-step explanation:

Let a represent the weight of Addie.

Let m represent the weight of Missy.

Let c represent the weight of Corky.

Together Addie and Missy weigh 18 ounces. This means that

a + m = 18 - - - - - - - - - 1

Missy and Corky weigh 22 ounces. This means that

m + c = 22

m = 22 - c - - - - - - - - - - 2

Addie and Corky weigh 12 ounces. This means that

a + c = 12

a = 12 - c - - - - - - - - - - - 3

Substituting equation 2 and equation 3 into equation 1, it becomes

22 - c + 12 - c = 18

34 - 2c = 18

- 2c = 18 - 34 = - 16

c = - 16/ - 2 = 8

Substituting c = 8 into equation 2, it becomes

m = 22 - 8

m = 14

Substituting c = 8 into equation 3, it becomes

a = 12 - 8

a = 4

Answer:

n>9

Step-by-step explanation:

4(n-3)-6>18

4n-12-6>18

4n-18>18

4n>18+18

4n>36

n>36/4

n>9

Answer:

d) [-16.03,-3.97]

[tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].

Step-by-step explanation:

Notation and previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

[tex]n_A=18[/tex] represent the sample of A

[tex]n_B =13[/tex] represent the sample of B

[tex]\bar x_A =39[/tex] represent the mean sample  for A

[tex]\bar x_B =49[/tex] represent the mean sample for B  

[tex]s_A =8[/tex] represent the sample deviation for A

[tex]s_B =4[/tex] represent the sample deviation for B

[tex]\alpha=0.01[/tex] represent the significance level

Confidence =99% or 0.99

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex] (1)  

The point of estimate for [tex]\mu_A -\mu_B[/tex] is just given by:  

[tex]\bar X_A -\bar X_B =39-49=-10[/tex]  

The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2=18+13-2=29[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.756[/tex]  

The standard error is given by the following formula:  

[tex]SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex]  

And replacing we have:  

[tex]SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188[/tex]  

Confidence interval  

Now we have everything in order to replace into formula (1):  

[tex]-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03[/tex]  

[tex]-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97[/tex]  

So on this case the 99% confidence interval for the differences of means would be given by [tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].

d) [-16.03,-3.97]

The 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], therefore the correct answer is F. None of the above.

This question is about computing a confidence interval for the difference of two sample means. The formula for the 99% confidence interval for the difference between two means is:

(X1 - X2) ± Z * sqrt [s1^2/n1 + s2^2/n2]

Where X1 and X2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. We plug the given values into the equation to calculate:

(39 - 49) ± 2.576 * sqrt [(8^2 / 18) + (4^2 / 13)] => -10 ± 2.576 * sqrt [3.56 + 1.23] => -10 ± 2.576 * sqrt [4.79] => -10 ± 2.576 * 2.19 => -10 ± 5.64

This means the 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], which is not among the given options, so the correct answer is F. None of the above.

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Answer: $49,875; $4,830C.

Step-by-step explanation:

The average salary of Teachers in a medium-sized suburban school district is $47,500 per year.

The standard deviation is $4,600

After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. The bonus of $500 will not alter the mean and standard deviation because equal amount is added for each teacher.

5% increase in each teacher's salary would differ. Therefore, it will affect the mean and standard deviation by 5%. Therefore, the new mean would be

47500 + (5/100 × 47500) = $49875

The new standard deviation would be

4600 + (5/100 × 4600) = $4830

Answer:

$50,375; $4,830 (I just answered it on one of my quizzes)

Step-by-step explanation:

Answer:

a) The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:

[tex]\hat p=\frac{348}{400}=0.87[/tex]

b) [tex]np=400*0.87=348>10[/tex]

[tex]n(1-p)=400(1-0.87)=52>10[/tex]

So both conditions are satisifed and we can construct the confidence interval.

c) The 95% confidence interval would be given (0.837;0.903).

We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903

d) On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.

e) The 99% confidence interval would be given (0.827;0.913).

We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913

f) The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.

Step-by-step explanation:

Data given and notation  

n=400 represent the random sample taken    

X=348 represent the number of graduates that found jobs in the sample

[tex]\hat p=\frac{348}{400}=0.87[/tex] estimated proportion of graduates that found jobs

[tex]\alpha[/tex] represent the significance level

z would represent the statistic to calculate the confidence interval

p= population proportion of graduates that found jobs

(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?

The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:

[tex]\hat p=\frac{348}{400}=0.87[/tex]

(b) Check if the conditions for constructing a confidence interval based on these data are met.

[tex]np=400*0.87=348>10[/tex]

[tex]n(1-p)=400(1-0.87)=52>10[/tex]

So both conditions are satisifed and we can construct the confidence interval.

(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.

The confidence interval would be given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.96[/tex]

And replacing into the confidence interval formula we got:

[tex]0.87 - 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.837[/tex]

[tex]0.87 + 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.903[/tex]

And the 95% confidence interval would be given (0.837;0.903).

We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903

(d) What does "95% confidence" mean?

On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.

(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data

For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=2.58[/tex]

And replacing into the confidence interval formula we got:

[tex]0.87 - 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.827[/tex]

[tex]0.87 + 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.913[/tex]

And the 99% confidence interval would be given (0.827;0.913).

We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913

(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider?

The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.

Answer:

Null hypothesis: The favorite sport of 53% of adults in the country is professional football ( with revenue of about $13 billion per year)

Alternate hypothesis: The adults in the country whose favorite sport is not professional football (with revenue of about $13 billion per year) is less than 53%

Step-by-step explanation:

The given claim is that the favorite sport of 53% of adults in the country is professional football (with revenue of about $13 per year) which is the null hypothesis. This means that the favorite sport of the remaining 47%(less than 53%) of adults in the country is not professional football (with revenue of about $13 billion per year) which is the alternate hypothesis