An individual is teaching a class on Excel Macros. The individual plans to break the class up into groups of 4 and wants each group to have 2 exercises to practice on, with no group doing the same exercise. The individual wants to know how many exercises he will need. Write an equation that expresses the situation, let x be the independent variable and y be the dependent variable

1. У-4x
2. y-2(x/4)
3. y-4x/2

The equation will be:

y = -2

Step-by-step explanation:

A horizontal line has no slope as it is parallel to x-axis.

The horizontal line is in the form y = b where b is the y-intercept.

Given

The line passes through (4 -2).

The y-coordinate of the given point is -2 which means that the line will intersect the y-axis on -2

So,

b = -2

and

The equation will be:

y = -2

Keywords: Equation of line, slope

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Answer:

Step 1

Step-by-step explanation:

Like terms are mathematical terms that have the exact same variables and exponents, this is why step 1 is the answer.

Answer : The correct option is, (B) Step 2

Step-by-step explanation :

The given expression is:

3(x - 2) = 3

In this expression, 'x' is a variable.

By using distributive property, we get:

3x - 6 = 3

Now adding 6 on both side, we get:

3x - 6 + 6 = 3 + 6

Now combining like terms, we get:

3x = 9

Now dividing the expression by 3, we get the value of 'x'.

x = 3

The meaning of like terms in mathematics is that have the same variables and exponents.

Hence, the step result of combining like terms is, Step 2

Answer:

b) {2, 7, 15, 31, 37}

Step-by-step explanation:

Ac is the complement of A, that is, the elements that are in the U(universe) but not in A.

Ac - {2,7,15,31}

[tex]B \cap C[/tex] are the elements that are in both B and C. So

(B ∩ C) = {15,37}

Ac U (B ∩ C) are the elements that are in at least one of Ac or (B ∩ C).

Ac U (B ∩ C) = {2,7,15,31,37}

So the correct answer is:

b) {2, 7, 15, 31, 37}

Answer:

Option b) is correct ie., [tex]A^{c}\bigcup (B \bigcap C)={\{2, 7, 15, 31, 37\}}[/tex]

Step-by-step explanation:

Given sets are

[tex]U ={\{2, 7, 10, 15, 22, 27, 31, 37, 45, 55\}}[/tex]

[tex]A = {\{10, 22, 27, 37, 45, 55\}}[/tex]

[tex]B = {\{2, 15, 31, 37\}}[/tex]

[tex]C = {\{7, 10, 15, 37\}}[/tex]

To find  [tex]A^{c}\bigcup (B \bigcap C)[/tex]

First to find [tex]A^{c}[/tex]

[tex]A^{c}={\{2,7,15,31\}}[/tex]

to find  [tex]B\cap C[/tex]

[tex]B\cap C={\{2, 15, 31, 37\}}\cap {\{7, 10, 15, 37\}}[/tex]

[tex]B\cap C={\{37,15\}}[/tex]

[tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31\}}\cup {\{37,15\}}[/tex]

[tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31,37\}}[/tex]

Therefore option b) is correct

Therefore  [tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31,37\}}[/tex]

Answer:

a) [tex]div F = 27 \frac{W}{km^3}[/tex]

b) [tex]\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}[/tex]

c) [tex]T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185[/tex]

Step-by-step explanation:

(a) Suppose that the actual heat generation is 27W/km3 What is the value of div F? div F- Include units)

For this case the value for div F correspond to the generation of heat.

[tex]div F = 27 \frac{W}{km^3}[/tex]

(b) Assume the heat flows outward symmetrically. Verify that [tex] F= \alpha r[/tex] where [tex]r=xi +yj+zk[/tex]. Find a α, (Include units.)

For this case we can satisfy this condition:

[tex]div[\alpha (xi +yj +z k)]]=\alpha(1+1+1)=3\alpha[/tex]

And since we have the value for the [tex]div F[/tex] we can find the value of [tex]\alpha[/tex] like this:

[tex]\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}[/tex]

(c) Let T (x,y,z) denote the temperature inside the earth. Heat flows according to the equation F= -k grad T where k is a constant. If T is in °C then k=27000 C/km. Assuming the earth is a sphere with radius 6400 km and surface temperature 20°C, what is the temperature at the center? 27,0 C/km.

For this case we have this:

[tex] F =-k grad T[/tex]

And grad T represent the direction of the greatest decrease related to the temperature.

So we have this equation:

[tex] 10(xi +yj+zk)=-27000 grad T[/tex]

And we can solve for grad T like this:

[tex] grad T = -\frac{10}{(27000)} (xi+yj+zk)[/tex]

Andif we integrate in order so remove the gradient on both sides we got:

[tex]T=-\frac{10}{2(27000)} (x^2 +y^2 +z^2) +C[/tex]

For our case we have the following condition:

[tex]x^2 +y^2 +z^2 = 6400 , T=20 C[/tex]

[tex] T=-\frac{1}{54000} (6400^2)+C =20[/tex]

And we can solve for C like this:

[tex] C= 20+\frac{6400^2}{5400}= 7605.185 [/tex]

So then our equation would be given by:

[tex]T=-\frac{10}{2(27000)} (x^2 +y^2 +z^2) +7605.185[/tex]

And for our case at the center we have that [tex]x^2+ y^2+ z^2 =0[/tex]

And we got:

[tex]T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185[/tex]

Answer:

b) ( N c ∩ D ) ∪ ( N ∩ D c )

Step-by-step explanation:

Let's see each option:

a) N ∪ D

This includes those who likes both, so this is not correc.t

b) ( N c ∩ D ) ∪ ( N ∩ D c )

Nc is the complement of N. That is those who do not subscribe to news magazine. Intensected with D, those are who do not subscribe to the news magazine but do to the daily informer. By the same logic, the second intersection are those who do subscribe to news magazine but not to the daily informer. This is the correct answer

c) N c ∩ D

This includes only those who subscribe to the daily informer but not to the news maganize. It also needs those who subscribe to the news magazine but not to the daily informer.

d) ( N ∩ D )c

This also involves those who do not subscribe to any of these news sources.

e) ( N c ∪ D ) ∩ ( N ∪ D c )

Wrong... Intersection of those who subscribe to the daily informer but not the news magazine and those who subscribe to the news magazine but not to the daily informer. This is the empty set.

Answer:

The product is positive, thus it is [tex]\bold{+7a^4b^3}[/tex]

Step-by-step explanation:

The full question in proper notation is:

"Find the product. If the result is negative, enter "-". If the result is positive, enter "+".

[tex]-7(-a^2)^2(-b^3)[/tex]"

We have to work with it using Order of operations know as well as PEMDAS, thus expression inside parenthesis go first and exponents.

On this expression we have to work with exponents

[tex](-a^2)^2 = (-a^2)(-a^2) =a^4[/tex]

Thus we get

[tex]-7(-a^2)^2(-b^3)=-7a^4(-b^3)[/tex]

Lastly we can work with multiplication and remembering that the multiplication of two negative signs becomes positive.

[tex]-7(-a^2)^2(-b^3)=7a^4b^3[/tex]

So the final simplified expression is [tex]\bold{7a^4b^3}[/tex]

Answer: +7a^4b^3

Step-by-step explanation:

Answer:

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Step-by-step explanation:

Notation and definitions  

n=50 represent the sample size  

[tex]\bar X= 27.2[/tex] represent the sample mean  

[tex]s=5.83[/tex] represent the sample standard deviation  

m represent the margin of error  

Confidence =88% or 0.88

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".  

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 88% of confidence, our significance level would be given by [tex]\alpha=1-0.88=0.12[/tex] and [tex]\alpha/2 =0.06[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=50-1=49[/tex]  

We can find the critical values in R using the following formulas:  

qt(0.06,49)

[1] -1.582366

qt(1-0.06,49)

[1] 1.582366

The critical value [tex]tc=\pm 1.582366[/tex]  

Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=1.582366 \frac{5.83}{\sqrt{50}}=14.613[/tex]  

With R we can do this:

m=1.582366*(5.83/sqrt(50))

m

[1] 1.304639

Calculate the confidence interval  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]27.02 - 1.582366 \frac{5.83}{\sqrt{50}}=25.715[/tex]  

[tex]27.02 + 1.582366 \frac{5.83}{\sqrt{50}}=28.325[/tex]

Using R the code is:

lower=27.02-m;lower

[1] 25.71536

upper=27.02+m;upper

[1] 28.32464

The 95% confidence interval is given by (25.71536 ;28.32464)  

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Answer:

There is a significant linear correlation between the two variables.

Step-by-step explanation:

Given that part time weekly earnings are

x y

10 93

15 171

20 204

20 156

35 261

Correlation 0.9199

H0: r=correlatin =0

Ha: r ≠0

(two tailed test)

r difference = 0.9199

t = [tex]r*\sqrt{(n-2)/(1-r^2)} \\=4.063[/tex]

p value < 0.0001

Since p <0.05 we reject null hypothesis

There is a significant linear correlation between the two variables.

This answer demonstrates how to create a scatter plot in Excel, interpret the correlation, calculate SSxx, SSyy, and SSxy, find t.05 from a t-distribution table, and perform a t-test to determine if the correlation coefficient is significantly different from zero.

In Excel, you can create a scatter plot by selecting the appropriate data and choosing Scatter Plot under the Insert tab. The scatter plot suggests a positive correlation between hours worked (X) and weekly pay (Y), indicating that as hours worked increase, so does weekly pay.

To calculate SSxx, SSyy, and SSxy, first, calculate the average of X and Y. Next, subtract the average from each individual data point and square the result to find SSxx and SSyy. For SSxy, subtract the average of X from each X data point and the average of Y from each Y data point, multiply these together, and take the sum.

With regards to finding t.05, look at the t-distribution table available with your study materials. Lastly, calculate the t test statistics using the formula t = r * sqrt((n-2)/(1-r^2)), where n is the number of pairs, r is the correlation coefficient obtained from your Excel scatter plot, and sqrt is the square root function. If the calculated t-value exceeds the t.05 value, you reject the null hypothesis (p = 0).

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Answer:

The pressure is changing at [tex]\frac{dP}{dt}=3.68[/tex]

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of [tex]\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}[/tex] and we want to find at what rate is the pressure changing.

The equation that model this situation is

[tex]PV^{1.4}=k[/tex]

Differentiate both sides with respect to time t.

[tex]\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\[/tex]

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

[tex]\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)[/tex]

Apply this rule to our expression we get

[tex]V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0[/tex]

Solve for [tex]\frac{dP}{dt}[/tex]

[tex]V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}[/tex]

when P = 23 kg/cm2, V = 35 cm3, and [tex]\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}[/tex] this becomes

[tex]\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68[/tex]

The pressure is changing at [tex]\frac{dP}{dt}=3.68[/tex].

Final answer:

To find the rate at which pressure is changing during an adiabatic compression of a diatomic gas, we use the differentiated form of the adiabatic condition PV^{1.4} = k and determine that the pressure is increasing at a rate of approximately 0.457 kg/(cm^2min).

Explanation:

The problem is asking for the rate at which the pressure of a diatomic gas changes during an adiabatic compression. Given the relationship PV^{1.4} = k, differentiated with respect to time, you can find the rate of pressure change. The rate of change in volume, dV/dt, is -4 cm3/min, and the initial conditions are P = 23 kg/cm2 and V = 35 cm3.

Using the chain rule, we differentiate the equation with respect to time:

d/dt (PV^{1.4}) = d/dt (k) => P * 1.4 * V^{0.4} * (dV/dt) + V^{1.4} * (dP/dt) = 0

When solved for the rate of pressure change dP/dt, this equation gives:

(dP/dt) = -P * 1.4 * V^{0.4} * (dV/dt) / V^{1.4}

Substituting the provided values into this equation yields:

(dP/dt) = -(23 kg/cm2) * 1.4 * (35 cm3)^{0.4} * (-4 cm3/min) / (35 cm3)^{1.4}

After calculation:

(dP/dt) ≈ 0.457 kg/(cm2min)

The pressure is increasing at a rate of approximately 0.457 kg/(cm2min).

Answer:

The 95% confidence interval would be given by [tex]-23.44 \leq \mu_1 -\mu_2 \leq -8.16[/tex]  

Step-by-step explanation:

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]\bar X_1 =114.1[/tex] represent the sample mean 1

[tex]\bar X_2 =129.9[/tex] represent the sample mean 2

n1=5 represent the sample 1 size  

n2=2 represent the sample 2 size  

[tex]s_1 =5.08[/tex] sample standard deviation for sample 1

[tex]s_2 =5.37[/tex] sample standard deviation for sample 2

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

Confidence interval

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

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =114.1-129.9=-15.8[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n_1 +n_2 -1=5+5-2=8[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that [tex]t_{\alpha/2}=\pm 2.31[/tex]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]

And replacing we have:

[tex]SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306[/tex]

Confidence interval

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

[tex]-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437[/tex]  

[tex]-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163[/tex]  

So on this case the 95% confidence interval would be given by [tex]-23.44 \leq \mu_1 -\mu_2 \leq -8.16[/tex]  

To calculate a 95% confidence interval for the difference between the true average stopping distances for cars equipped with system 1 and system 2, use the formula CI = (x1 - x2) ± (c * SE) where x1 and x2 are the sample means, and SE is the standard error of the difference.

To calculate a 95% confidence interval for the difference between the true average stopping distances for cars equipped with system 1 and system 2, we can use the formula:


CI = (x1 - x2) ± (c * SE)


where x1 and x2 are the sample means, and SE is the standard error of the difference. The critical value c can be found using the t-distribution table for the given sample sizes n1 and n2.

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Answer:

The value of the test statistic z for this hypothesis test is -2.79

Step-by-step explanation:

Consider the provided information.

To calculate the test statistic use the formula:

[tex]z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}[/tex]

Where, z  is Test statistics, n is Sample size, p₀ = Null hypothesized value  and [tex]\hat p[/tex] = Observed proportion.

p₀ = 0.50

Thus 1-p₀= 0.50

42% responded that if they had a cell phone, thus [tex]\hat p=0.42[/tex]

The sample size is 305.

Substitute the respective values in the above formula.

[tex]z=\frac{0.42-0.50}{\sqrt{\frac{0.50(0.50)}{305}}}[/tex]

[tex]z=\frac{-0.08}{\sqrt{0.00082}}[/tex]

[tex]z=-2.79[/tex]

Hence, the value of the test statistic z for this hypothesis test is -2.79

The value of the test statistic [tex]\( z \)[/tex] for this hypothesis test is approximately -2.797.

To calculate the test statistic [tex]\( z \)[/tex], we follow these steps:

1. State the null hypothesis [tex]\( H_0: p = 0.50 \)[/tex] and the alternative hypothesis [tex]\( H_A: p < 0.50 \)[/tex].2. Identify the sample proportion [tex]\( \hat{p} \)[/tex], which is the proportion of the sample that has the characteristic of interest.

In this case, [tex]\( \hat{p} = \frac{128}{305} \approx 0.42 \)[/tex]

3. Calculate the standard error of the proportion, which is given by [tex]\( SE = \sqrt{\frac{p(1-p)}{n}} \)[/tex], where [tex]\( p \)[/tex] is the population proportion under the null hypothesis and [tex]\( n \)[/tex] is the sample size

[tex]\( SE = \sqrt{\frac{0.50(1-0.50)}{305}} \approx \sqrt{\frac{0.25}{305}} \)[/tex]

4. Compute the test statistic [tex]\( z \)[/tex] using the formula [tex]\( z = \frac{\hat{p} - p}{SE} \)[/tex], where [tex]\( \hat{p} \)[/tex] is the sample proportion, [tex]\( p \)[/tex] is the population proportion under the null hypothesis, and [tex]\( SE \)[/tex] is the standard error.

5. Plug in the values to get [tex]\( z = \frac{0.42 - 0.50}{\sqrt{\frac{0.25}{305}}} \)[/tex].

6. Simplify the expression to find the value of [tex]\( z \)[/tex].

[tex]\[ z = \frac{0.42 - 0.50}{\sqrt{\frac{0.25}{305}}} = \frac{-0.08}{\sqrt{\frac{1}{1220}}} = \frac{-0.08}{\sqrt{0.0008197}} = \frac{-0.08}{0.0286} \approx -2.797 \][/tex]

To summarize the results, approximately 51.5% of the qualified applicants were accepted into the magnet school programs, around 14.5% were waitlisted, and about 34% were turned away due to lack of space.

Given that;

An article reported on a school​ district's magnet school programs.

Of the 18701870 qualified​ applicants, 963963 were​ accepted, 271271 were​ waitlisted, and 636636 were turned away for lack of space.

Let's calculate the relative frequency for each decision made.

To find the relative frequency, we divide the number of applicants by the total number of qualified applicants.

For the accepted applicants:

Relative frequency = Number of accepted applicants / Total qualified applicants

Relative frequency = 963 / 1870

Relative frequency ≈ 0.515

For the waitlisted applicants:

Relative frequency = Number of waitlisted applicants / Total qualified applicants

Relative frequency = 271 / 1870

Relative frequency ≈ 0.145

For the applicants turned away:

Relative frequency = Number of turned away applicants / Total qualified applicants

Relative frequency = 636 / 1870

Relative frequency ≈ 0.340

Thus, To summarize the results, approximately 51.5% of the qualified applicants were accepted into the magnet school programs, around 14.5% were waitlisted, and about 34% were turned away due to lack of space.

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Answer:

[tex] t=(15-1) [\frac{0.219}{0.2}]^2 =16.7864[/tex]

[tex]p_v = 2*P(\chi^2_{14}>16.786)=0.5355[/tex]

And the 2 is because we are conducting a bilateral test.

[tex]\alpha=0.05[/tex] since the the [tex]p_v >0.05[/tex] we fail to reject the null hypothesis.

[tex]\alpha=0.01[/tex] since the the [tex]p_v >0.01[/tex] we fail to reject the null hypothesis.

Step-by-step explanation:

Data given

[tex]\mu=7[/tex] population mean (variable of interest)

[tex]\sigma=0.2[/tex] represent the population standard deviation

[tex]s=0.219[/tex] represent the sample deviation

n=15 represent the sample size

[tex]\alpha=0.05,0.01[/tex] represent the values for the significance level  

Hypothesis test

On this case we want to check if the population standard deviation is equal or not to 0.2, so the system of hypothesis are:

H0: [tex]\sigma = 0.2[/tex]

H1: [tex]\sigma \neq 0.2[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(15-1) [\frac{0.219}{0.2}]^2 =16.7864[/tex]

P value

We know the degrees of freedom of the distribution 14 on this case and we can find the p value like this:

[tex]p_v = 2*P(\chi^2_{14}>16.786)=0.5355[/tex]

And the 2 is because we are conducting a bilateral test.

[tex]\alpha=0.05[/tex] since the the [tex]p_v >0.05[/tex] we fail to reject the null hypothesis.

[tex]\alpha=0.01[/tex] since the the [tex]p_v >0.01[/tex] we fail to reject the null hypothesis.

The question involves hypothesis testing for variance in relation to a coffee vending machine. After calculating the test statistic (using the Chi-square test) and comparing it with critical values at .05 and .01 significance levels, we fail to reject the null hypothesis. This means that there's not enough statistical evidence to suggest that the new machine has different variability in coffee amounts dispensed compared to the old machine.

To solve this problem, you need to use hypothesis testing for variance. Hypothesis testing is a statistical method that is used in making statistical decisions using experimental data. In hypothesis testing, an initial claim or belief about a population is translated into two competing hypotheses, the null hypothesis and the alternative hypothesis.

In this case, the null hypothesis(H0) is that the variance σ² is equal to 0.04 (σ=0.2; σ^2 = (0.2)² = 0.04), and the alternative hypothesis(Ha) is that the variance is not equal to .04.

The test statistic for this hypothesis is the Chi-Square statistic. It's calculated using the formula:

[ (n-1)*s² ] / σ²

where n is the sample size, σ² is the variance under the null hypothesis and s² is the empirical or sample variance. In your case, n = 15, σ² = 0.04 and s² = (0.219)² = 0.047961.

Substituting the values, we get (chi-square) χ² = (15-1)*(0.047961)/0.04 = 0.5995125.

The degrees of freedom (df) in this context is (n-1) = 14. At .05 level of significance, for 14 degrees of freedom, the critical values of χ² are 6.571 and 23.684. Likewise, for .01 level of significance, critical values are 3.787 and 30.578.

The calculated statistic (0.5995125) falls between these ranges for both cases, so we fail to reject the null hypothesis at both .05 and .01 significance level. Hence, the manufacturer does not have enough evidence to suggest that the new machine has a different variability in the amount of coffee dispensed compared to the old machine.

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Answer:

a) 92% Confidence interval: (1027.5,1048.5)

b) Sample size = 100

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 1038

Sample size, n = 25

Alpha, α = 0.08

Population standard deviation, σ = 30

a) 92% Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.08} = \pm 1.75[/tex]

[tex]1038 \pm 1.75(\frac{30}{\sqrt{25}} ) = 1038 \pm 10.5 = (1027.5,1048.5)[/tex]

b) In order to reduce the confidence interval by half, we have to quadruple the sample size.

Thus,

[tex]\text{Sample size} = 25\times 4 = 100[/tex]

A 92% confidence interval for the mean lifetime of a 60-watt bulb is (1027.5, 1048.5) hours. To halve this interval while maintaining the same confidence level, the sample size would need to be increased to 100.

The lifetime of a light bulb, expressed as a random variable, is said to have a Normal distribution with σ = 30 hours. The given random sample consists of 25 bulbs with a sample mean lifetime of = 1038. To construct a 92% confidence interval estimate for the mean lifetime (μ), we first need to identify the standard error, which is σ/√n => 30/√25 = 6. To calculate the confidence interval, we adjust the sample mean by a few standard errors. For a 92% confidence interval, the Z score is approximately ±1.75 (obtained from a Z distribution table). Therefore, the interval is 1038 ± 1.75 * 6 = 1038 ± 10.5. Hence, the 92% confidence interval for the population mean is (1027.5, 1048.5). To halve the confidence interval at the same confidence level (i.e. to make it ±5.25), we need to halve the standard error. Since the standard error is inversely proportional to the square root of the sample size, we need to quadruple the sample size to halve it. So, a sample size of 100 would be required.

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Answer:

Option d) is correct

ie, quotient property

Step-by-step explanation:

Given expression is log 618-log 66=log 63

Now we take log 618-log 66

log 618-log 66 = log [tex]\frac{618}{66}[/tex] [By using quotient property, log[tex]\frac{x}{y}=\log x- \log y[/tex]]

= log 63

Therefore log 618- log 66= log 63

Option d) is correct

In the given expression we are using the quotient property

Answer:

10000

Step-by-step explanation:

Answer:

[tex]z=\frac{0.41-0.43}{\sqrt{\frac{0.41(1-0.41)}{700}+\frac{0.43(1-0.43)}{200}}}=-0.51[/tex]    

[tex]p_v =2*P(Z<-0.51)=0.614[/tex]  

And we can use the following R code to find it: "2*pnorm(-0.505)"

The p value is a very high value and using any significance given [tex]\alpha=0.05[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.  

Step-by-step explanation:

1) Data given and notation  

[tex]n_{1}=700[/tex] sample of young adults (age 19 to 35)

[tex]n_{2}=200[/tex] sample of children age 19 to 35

[tex]p_{1}=0.41[/tex] represent the proportion of young adults said they thought parents would provide financial support

[tex]p_{2}=0.43[/tex] represent the proportion of parents said they would provide support

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic (variable of interest)  

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for the two samples are different , the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} = p_{2}[/tex]  

Alternative hypothesis:[tex]p_{1} \neq p_{2}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\frac{p_1 (1-p_1)}{n_{1}}+\frac{p_2 (1-p_2)}{n_{2}}}}[/tex]   (1)  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.41-0.43}{\sqrt{\frac{0.41(1-0.41)}{700}+\frac{0.43(1-0.43)}{200}}}=-0.51[/tex]    

4) Statistical decision

We can calculate the p value for this test.    

Since is a two tailed  test the p value would be:  

[tex]p_v =2*P(Z<-0.51)=0.614[/tex]  

And we can use the following R code to find it: "2*pnorm(-0.505)"

The p value is a very high value and using any significance given [tex]\alpha=0.05[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.  

Answer:

(a) z-value = -2.12

(b) p-value = 0.0170

(c) Reject H0

Step-by-step explanation:

(a)

[tex]std-err=\frac{std-dev}{\sqrt{n}}=\frac{2}{\sqrt{50}}=0.2828[/tex]

[tex]z-value=\frac{X-mean}{std-err}=\frac{19.4-20}{0.2828}=-2.12[/tex]

(b)

with

z-value = -2.12

significance level = 0.05

one-tail hypothesis (H0: μ ≥ 20)

We can see on the normal distribution table (Z-Score table) that

p-value = 0.0170

(c)

Since p value (0.0170) is less than α=0.05 we reject H0

Hope this helps!

Answer:

[tex]z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467[/tex]  

[tex]p_v =P(Z>-2.467)=0.0068[/tex]  

So the p value obtained was a very low value and using the significance level assumed for example [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of circuits that show evidence of undercutting is significantly less than 0.05.  

Step-by-step explanation:

1) Data given and notation  

n=1000 represent the random sample taken

X=33 represent the number of circuits that show evidence of undercutting

[tex]\hat p=\frac{33}{1000}=0.033[/tex] estimated proportion of circuits that show evidence of undercutting

[tex]p_o=0.05[/tex] is the value that we want to test

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

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that they reduce the rate of undercutting to less than 5%.:  

Null hypothesis:[tex]p\geq 0.05[/tex]  

Alternative hypothesis:[tex]p < 0.05[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(Z>-2.467)=0.0068[/tex]  

So the p value obtained was a very low value and using the significance level assumed for example [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of circuits that show evidence of undercutting is significantly less than 0.05.  

Answer:

Age    |    Believe   |   Not believe  |   Total

<45     |    0.148      |        0.422       |   0.570

>45     |    0.301      |        0.129        |   0.430

Step-by-step explanation:

We have to construct a probability matrix for this problem.

Of the people surveyed, 57% were under age 45. That means that 43% is over age 45.

70% of the ones who were 45 or older, believe the Social Security system will be secure in 20 years.

The Believe proportion is 51%.

Then, the proportion that believe and are under age 45 is:

[tex]0.51=P(B;<45)*0.43+0.70*0.57\\\\P(B;<45)=\frac{0.51-0.70*0.57}{0.43} =\frac{0.11}{0.43}= 0.26[/tex]

We can now construct the probability matrix for one respondant selected randomly:

[tex]P(<45\&B)=0.57*0.26=0.148\\\\P(<45\&NB)=0.57*(1-0.26)=0.57*0.74=0.4218\\\\P(>45\&B)=0.43*0.7=0.301\\\\P(>45\&NB)=0.43*(1-0.7)=0.43=0.3=0.129[/tex]

Age    |    Believe   |   Not believe  |   Total

<45     |    0.148      |        0.422       |   0.570

>45     |    0.301      |        0.129        |   0.430

1. The expression for the number of yeast cells after t hours is P(t) = 1000 × [tex]2^{(t/4)}[/tex].

2. After 10 hours, there will be 4000 yeast cells.

3. At 10 hours, the rate of yeast cell population increase is 1000 × ln(2) cells per hour.

1. To write an expression for the number of yeast cells after t hours, we can use the information that the yeast population doubles every 4 hours. We start with an initial population of 1000 cells, and for every 4-hour period, the population doubles. Therefore, the expression can be written as:

P(t) = 1000  ×[tex]2^{(t/4)}[/tex]

Where P(t) represents the number of yeast cells after t hours.

2. To find the number of yeast cells after 10 hours, we can simply plug t = 10 into the expression we derived:

P(10) = 1000 × [tex]2^{(10/4)}[/tex]

P(10) = 1000 × [tex]2^{(2)}[/tex]

P(10) = 1000 × 4

P(10) = 4000

So, there will be 4000 yeast cells after 10 hours.

3. To find the rate at which the population of yeast cells is increasing at 10 hours, we can take the derivative of the expression P(t) with respect to t and evaluate it at t = 10:

P'(t) = (1000/4) × [tex]2^{(t/4)}[/tex] × ln(2)

P'(10) = (1000/4) × [tex]2^{(10/4)}[/tex] × ln(2)

P'(10) = (1000/4) × [tex]2^{(2)}[/tex] × ln(2)

P'(10) = 250 × 4  ln(2)

P'(10) = 1000 × ln(2)

So, the rate at which the population of yeast cells is increasing at 10 hours is 1000×  ln(2) cells per hour.

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Complete question below :

1. Write an expression for the number of yeast cells after t hours.

2. Find the number of yeast cells after 10 hours.

3. Find the rate at which the population of yeast cells is increasing at 10 hours.

An exponential growth model for yeast growth gives P(t)=1000e^(ln(2)t/4). After 10 hours, the yeast population is approximately 5657 cells. The rate of increase at 10 hours is about 978 cells per hour.

Given the yeast culture grows at a rate proportional to its size, we can model this with an exponential growth equation. The general form is:

P(t) = P0ekt

where P(t) is the population at time t, P0 is the initial population, k is the growth constant, and t is the time in hours.

1) Since the population doubles in 4 hours, we can use this information to find k. Start with the equation when the population doubles:

P(4) = 2P0

Substitute P0 and t = 4:

2P0 = P0e4k

Divide both sides by P0:

2 = e4k

Take the natural logarithm of both sides:

ln(2) = 4k

Solve for k:

k = ln(2) / 4

Now substitute k back into the general equation:

P(t) = 1000e(ln(2)t/4)

2) To find the number of yeast cells after 10 hours, substitute t = 10:

P(10) = 1000e(ln(2)10/4)

Simplify the exponent:

P(10) = 1000e(2.5ln(2))

P(10) = 1000 * 22.5

P(10) ≈ 1000 * 5.657

P(10) ≈ 5657 cells

3) To find the rate of increase at 10 hours, we need to differentiate P(t):

dP/dt = 1000 * (ln(2) / 4) * e(ln(2)t/4)

Substitute t = 10:

dP/dt = 1000 * (ln(2) / 4) * 2(10/4)

dP/dt = 1000 * (ln(2) / 4) * 2.5

dP/dt ≈ 1000 * 0.173 * 5.657

dP/dt ≈ 978 cells per hour

Complete question below :

A culture of yeast grows at a rate proportional to its size. If the initial population is 1000 cells and it doubles after 4 hours, answer the following questions:

1. Write an expression for the number of yeast cells after t hours.

2. Find the number of yeast cells after 10 hours.

3. Find the rate at which the population of yeast cells is increasing at 10 hours.

Answer:

The Normal approximation (with continuity correction) of the probability P(13 < X ≤ 16) is equal to: P(13.5<X<16.5).

Step-by-step explanation:

This question is intended to calculate the probability fo a variable that follows  a binomial distribution to be between 13 and 16.

When approximated to a Normal distribution, a correction for continuity has to be made, because the binomial distribution is a discrete value function and the normal function is a continous value function.

For X>13, it should be rewritten as X>13.5 (it substracts because it does not include 13).

For X≤16, it should be rewritten as X<16.5 (it adds because it includes 16).

The Normal approximation (with continuity correction) of the probability P(13 < X ≤ 16) is equal to: P(13.5<X<16.5).

Answer: 0.0479

Step-by-step explanation:

Given : Computer Help Hot Line receives, on average, 14 calls per hour asking for assistance.

Let x be number of variable that denotes the number of calls that follows a Poisson distribution.

Poisson distribution formula : [tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]

, where [tex]\lambda[/tex] =Mean of the distribution.

Here ,

Then, the probability that the company will receive more than 20 calls per hour= [tex]P(x>20)=1-P(x\leq20)[/tex]

[tex]=1-0.9521=0.0479 [/tex]  

(From Cumulative Poisson distribution table the value of P(x ≤ 20) =0.9521 corresponding to  [tex]\lambda=14[/tex] ).

Thus , the probability that the company will receive more than 20 calls per hour = 0.0479

Answer:

c. standard deviation, median, range

Step-by-step explanation:

The standard deviation without the Bessel's correct is defined as:

[tex] s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n}[/tex]

And if we find the expected value for s we got:

[tex] E(s^2) = \frac{1}{n} \sum_{i=1}^n E(x_i -\bar x)^2 [/tex]

[tex] E(s^2)= \frac{1}{n} E[\sum_{i=1}^n ((x_i -\mu)-(\bar x -\mu)^2)][/tex]

We have this:

[tex] E(\sum_{i=1}^n(x_i-\mu)^2) =n\sigma^2[/tex]

[tex]E[\sum_{i=1}^n (x_i -\mu)(\bar x -\mu)]= \sigma^2[/tex]

[tex]E[\sum_{i=1}^n (\bar x -\mu)^2]=\sigma^2[/tex]

[tex]E(s^2)=\frac{1}{n} (n\sigma^2 -2\sigma^2 +\sigma^2)[/tex]

[tex]E(s^2)=\frac{n-1}{n}\sigma^2[/tex]

as we can see the sample variance is a biased estimator since:

[tex]E(s^2)\neq \sigma^2[/tex]

And we see that the standard deviation is biased, since:

[tex] E(s) = \sqrt{\frac{n-1}{n}} \sigma[/tex]

because [tex]E(s)\neq \sigma[/tex]

The mean is not biased for this case option a is FALSE.

The proportion is not biased for this reason option d is FALSE

The range can be considered as biased since we don't have info to conclude that the range follows a distirbution in specific.

The sample median "is an unbiased estimator of the population median when the population is normal. However, for a general population it is not true that the sample median is an unbiased estimator of the population median".

And for this reason the best option is c.

Answer: standard deviation, median, range

Step-by-step explanation:

Answer: The correct option is (b)

Using a single dummy variable in a regression model

Step-by-step explanation:

A regression model is a model that measures the relationship between a dependent variable and one or more independent variables. In this question the dependent variable y is the sales of the car and the independent variable is the choice of the gender of a salesperson, Which is single dummy variable.

Answer:

Step-by-step explanation:

As the question is not complete, we will generalize the statement as follows:

Olanda needs "A" for a future project. She can invest "x" now at an annual rate of "R" , compounded monthly. Assuming that no withdrawals are made, how long will it take for her to have enough money for her project?

A= Amount Olanda needs (future value).

X= Amount available for Olanda.

R= Anual rate.

First, we have that the future value is given by:

[tex]A=x(1+\frac{R}{n}) ^{nt}[/tex]

Where n=12 because the rate is annual.

And we need to calculate the time it will take with that annual rate to get the money she needs (future value) from what she has now (present value). So, we must make it explicit for "t"; solving:

We have:

[tex]\frac{A}{x} = (1+ \frac{R}{n})^{nt} [/tex]

[tex]ln[\frac{A}{x}]=ln[ (1+ \frac{R}{n})^{nt}][/tex]

Applying logarithmic properties

[tex]ln[\frac{A}{x}]=nt*ln[1+\frac{R}{n}][/tex]

Finally, we have:

Answer:

17x² - 5x + 11 = 0

Step-by-step explanation:

To add or subtract a polynomial do the operation with like - terms.

i.e., Two terms of x² gets added or subtracted. Similarly terms of x and constant.

Here, we have to subtract:

[tex]$ 2x^2 + 9x - 3 - 7x^2 + 4x - 8 - 9x^2 - 13x + 5 - 9x^2 + 13x - 5 + 5x^2 - 13x - 1 + 5x^2 + 5x - 5 $[/tex]

Club all the like terms for easier simplification. We get:

[tex]$ (-2 - 7 - 9 - 9 + 5 + 5)x^2 + (9 + 4 - 13 + 13 - 13 + 5)x + (-3 - 2 + 5 - 5 - 1) $[/tex]

[tex]$ \implies - 17 x^2 + 5x - 11 = 0 $[/tex]

Multiplying by -1 throughout:

17x² - 5x + 11 = 0 is the answer.

Answer:

Option e) 0.10 < P < 0.15

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $45,000

Sample mean, [tex]\bar{x}[/tex] = $44,500

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = $1,750

First, we design the null and the alternate hypothesis

[tex]H_{0}: m = 44500\\H_A: m < 44500[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{44500 - 45000}{\frac{1750}{\sqrt{20}} } = -1.2778[/tex]

Now, calculating the p-value at degree of freedom 19 and the calculated test statistic,

p-value = 0.108494

Thus,

Option e) 0.10 < P < 0.15

Answer:there will be 211 reflectors on 1 mile of highway

Step-by-step explanation:

A mile is equal to 5280 feet. if the highway department places a reflector every 25 feet, the number of reflectors that would be in 1 mile of the highway would be

Total number of feets / 25. It becomes. 5280/25 = 211.2 reflectors. Approximating 211.2, it becomes 211 reflectors.

Answer:

distance is maximum at coordinates (−3, 0) , (3, 0) and minimum at distance (0,-9)

Step-by-step explanation:

since the distance to the statue is

D² = (x-x₀)²+ (y-y₀)²

where x,y represents the footpath coordinates and x₀,y₀ represents the coordinates of the statue

and

y= x²-9   , for x  [−3, 3]

x² = y+9

thus

D² = x²+ y²

D² = y+9 +y²

since D² is minimised when d is minimised, then

the change in distance with y is

d (D²)/dy =  2*D*d(D)/dy =2*D*( 1+2*y)

d (D²)/dy =2*D*( 1+2*y)

since D>0 , d (D²)/dy >0 for y> -1/2

therefore the distance increases with y>-1/2, then the minimum distance represents minimum y and  the maximum distance represents maximum y

since

y= x²-9  for [−3, 3]

y is maximum at x=−3 and x=3 → y=0

and minimum for x=0   → y=-9

then

distance is maximum at coordinates (−3, 0) , (3, 0) and minimum at distance (0,-9)