System of Linear Equations In Exercises 25–38, solve the system using ei ther Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Answer:

Step-by-step explanation:

Given that price, p, and quantity, x, are inearly related

We are given two points on this line as (p,x) = (96.90, 121) and (40.20,310)

Using two point formula we find linear equation as

[tex]\frac{y-121}{310-121} =\frac{0-96.90}{40.20-96.90} \\(x-121)(-56.70)=189(p-96,.90)\\189p+56.70x =25174.80[/tex]

a)[tex]189p+56.70x =25174.80[/tex] is the linear equation

b) A(x)

Substitute to get

189(50.70)+56.70x = 25174.80

x=275 units

c) This is linear function hence no local maxima or minima

d) No maximia or minima

Answer:

0.0625

Step-by-step explanation:

The prevalence of gene a = 25 %, P (a) = 0.25

birth defect occurs when both parents have prevalence of gene a.

P (Defect) = P ( Both parents have gene a)

If both parents inherit the gene a independently, the the individual will have a birth defect when both parents have gene a.

P ( Father having gene a) = 0.25

P ( Mother having gene a) = 0.25

Hence,

P (Birth Defect) = P ( Both parents have gene a) = 0.25 * 0.25 = 0.0625

Answer:

At least two of the restaurant have different mean delivery time.

Step-by-step explanation:

The ANOVA known as Analysis of variance involves the hypothesis testing of means on the basis of variances. The null hypothesis in ANOVA is taken as the equality mean i.e. H0: All means are equal. Whereas alternative hypothesis consists of at least two means are not equal.

The problem states that a pizza lover is comparing average delivery time. The null and alternative hypothesis in this case would be

H0: All restaurant have equal mean delivery time.

H1: At least two restaurant have different mean delivery time.

In an ANOVA comparing four local pizza restaurants' delivery times, the alternative hypothesis is that at least two restaurants have different mean delivery times.

When performing an ANOVA with data comparing the average delivery times for four local pizza restaurants, the alternative hypothesis is that at least two of the restaurants have different mean delivery times. This is because ANOVA tests the null hypothesis that all group means are equal against the alternative that at least one group mean is different. Specifically, the null hypothesis for a one-way ANOVA test with four groups is μ1 = μ2 = μ3 = μ4, where each μ represents the mean delivery time of a different restaurant. The alternative hypothesis, which the student is inquiring about, is Ha: μi ≠ μj for some i ≠ j.

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Answer:

The question is not so clear and complete

Step-by-step explanation:

But for questions like this, since the equation has been given, what is expected is for us to make comparison, compare the RHS with the LHS or by method of comparing coefficients.

We follow the stated conditions since we are told that b and c are both integers which are greater than 1 and b is less than the product of cb. from these conditions, we can compare and get the values of b , c and d.

Another approach is to assume values, make assumptions with the stated conditions, however, our assumptions must be valid and correct if we substitute the assumed values of b, c and d in the equation, it must arrive at the same answer for the RHS. i.e LHS = RHS

Answer: m1 = 4

m2 = 5

m3 = 2

Step-by-step explanation:

given (21/11, 6/11) = m1 (-1/3) + m2 (3, -2) + m3 (5, 2)

= (-m1 + 3m2 + 5m3) / 11 = 21/11

= (3m1 + (-2)m2 + 2m3) / 11 = 6/11

so that m1 + m2 +m3 = 11

-m1 + 3m2 + 5m3 = 21

3m1 - 2m2 + 2m3 = 6

from this, we get the augmented matrix as

\left[\begin{array}{cccc}-1&1&1&11\\-1&3&5&21\\3&-2&2&6\end{array}\right]

= \left[\begin{array{cccc}-1&1&1&11\\0&4&6&32\\0&-5&-1&-27\end{array}\right]  \left \{ {{R2=R2 + R1} \atop {R3=R3 -3R1 }]} \right.

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&-5&-1&-27\end{array}\right]

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&0&13/2&13\end{array}\right]

(R3 = R3 + 5R2)

this gives m1 + m2 + m3 = 11

m2 + 3/2 m3 = 8

13/2 m3 = 8

13/2 m3 = 13

m3 = 2

m2 = 8 -3/2 (2) = 5

= m1 = 11- 5 - 2 = 4

this gives

m1 = 4

m2 = 5

m3 = 2

To achieve the given center of mass, 4 kg of mass should be allocated to the vector u1, 6 kg to the vector u2, and 1 kg to the vector u3.

The provided equation for the center of mass expresses a weighted average of the position vectors, with each mass acting as the weight on its respective vector. We are given three vectors and a total mass, and must find the weights to apply to each vector in order to achieve a particular center of mass. Setting up equations for each component of the center of mass, we get two equations:

21/11 = (-1m1 + 3m2 + 5m3) / 11 and 6/11 = (3m1 - 2m2 + 2m3) / 11.

These equations can then be solved simultaneously to determine the values of m1, m2, and m3. Solving these equations gives m1 = 4 kg, m2 = 6 kg, and m3 = 1 kg. So, four kilograms of mass should be allocated to the vector u1, six kilograms to the vector u2, and one kilogram to the vector u3.

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If events A and B are mutually exclusive, the probability of A or B occurring is equal to the sum of the probabilities of A and B. Given that P(A or B) = 0.5 and P(B) = 0.3, P(A) can be calculated as 0.2.

If events A and B are mutually exclusive, the probability of A or B occurring is equal to the sum of the probabilities of A and B. So, we have P(A or B) = P(A) + P(B). Given that P(A or B) = 0.5 and P(B) = 0.3, we can substitute these values into the formula to solve for P(A).

0.5 = P(A) + 0.3.

Now, subtract 0.3 from both sides to isolate P(A):

0.5 - 0.3 = P(A).

P(A) = 0.2.

Therefore, the probability of event A occurring is 0.2.

Answer:

break even point = 8000 socks produced or $36000 in costs

Step-by-step explanation:

the cost function of the firm is

total cost = fixed cost + variable cost = $20000 + $2*Q

where Q= number of socks

the revenue from sales is

sales = Price* Q = $4.50*Q

the break even point is reached when the net profit is = 0 ( that is, the total cost is equal to the revenue from sales) , then

total cost = sales

$20000 + $2*Q =$4.50*Q

Q= $20000/($4.50-$2) = 8000 socks

that represents

total cost = $20000 + $2*8000  = $36000

then

break even point = 8000 socks produced or $36000 in costs

Answer: N >/= 7 bits

Minimum of 7 bits

Step-by-step explanation:

The minimum binary bits needed to represent 65 can be derived by converting 65 to binary numbers and counting the number of binary digits.

See conversation in the attachment.

65 = 1000001₂

65 = 7 bits :( 0 to 2^7 -1)

The number of binary digits is 7

N >/= 7 bits

To represent the unsigned decimal integer 65 in binary format, we need a minimum of 7 binary bits.

This question is related to the conversion of decimal numbers to binary format. To present an unsigned decimal 65 in binary, we first need to find the highest power of 2 that is less than or equal to 65, which is 2^6=64.

Then we continue to find the next highest power of 2 for the remainder of the value until we reach zero. This process would require a total of 7 bits. Therefore, to represent 65 as an unsigned binary integer, we need a minimum of 7 bits.

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In this problem you will calculate ∫302+4 by using the formal definition of the definite integral: ∫()=lim→∞[∑=1(∗)Δ].

(a) The interval [0,3] is divided into equal subintervals of length Δ. What is Δ (in terms of )? Δ =

(b) The right-hand endpoint of the th subinterval is denoted ∗. What is ∗ (in terms of and )? ∗ =

Answer:

a) Δ= [tex]\frac{3}{n}[/tex]

b) [tex]x^{*}_{k} = \frac{3k}{n}[/tex]

Step-by-step explanation:

a)  If the interval [0,3] , i.e let a = 0 , b =3 and n=n.

So [0,3] divide into n equal subintervals;

Therefore, the length Δ= [tex]\frac{b-a}{n}[/tex]

Δ= [tex]\frac{3-0}{n}[/tex]

Δ= [tex]\frac{3}{n}[/tex]

b) To calculate [tex]x^{*}_{k}[/tex];

[tex]x^{*}_{k}[/tex] = a + k . Δ          (where n= 0, Δ = [tex]\frac{3}{n}[/tex])

= 0 + k . [tex]\frac{3}{n}[/tex]

[tex]x^{*}_{k}[/tex]  =  [tex]\frac{3}{k}[/tex]

The right-hand endpoint of the kth subinterval, denoted x∗k, can be expressed as a function of both k and n (the total number of subintervals). The formula x∗k = k/n can be used in this context, where k denotes the position of the subinterval and n represents the total number of subintervals.

In the context of subintervals, x∗k represents the right-hand endpoint of the kth subinterval. When a range is divided into n subintervals, the right-hand endpoint of the kth subinterval could be represented as a function of both k and n. Thus, x∗k can be expressed as k/n.

To illustrate, suppose we have a range from 0 to 1 (inclusive) and want to split it into 4 subintervals (n=4). The right endpoint of the 1st subinterval (k=1) would be 1/4 = 0.25. For the 2nd subinterval (k=2), the endpoint would be 2/4 = 0.5. And so on.

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Answer: x =6 1/3

Step-by-step explanation:

Whichever value of x from the given option that make f(x) = 0 is the zero of f(x). x = 6 1/3 satisfies this, you can check by replacing 6 1/3 by x in the equation. 6 1/3 can be written as 19/3

f(19/3) = 9(19/3)² - 54(19/3) - 19

= 9(361/9) - 54(19/3) - 19

= 361 - 342 - 19

= 0.

The remaining values don't give 0.

Answer:

The probability that we have a fair coin is 0.2717

Step-by-step explanation:

Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)

We also let H be the event that the coin which is flipped lands (which is head)

Therefore, the problem arise:

That:  P(H/C1) = 1/2.............fair coin

          p(H/C2) = 2/3..........Chance of head

          p(H/C3) = 2/3............Chance of tail

Now, noting that the coin was picked at random,

We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3

We can then say that or calculate thus:

P( C2 | H ) = P (H ∩ C2) ÷ P(H)

Where P (H ∩ C2) means the probability of event intersection

= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)

= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)

= (1 / 2) (1 / 3) ÷  (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)

0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)

0.15 ÷ 0.15 + 0.201 + 0.201

=0.15 / 0.552

= 0.2717

Answer:

[tex]r=\sqrt{6/sinucosu}[/tex]

Step-by-step explanation:

To convert to polar form:

[tex]x=rcosu[/tex]

[tex]y=rsinu[/tex]

We substitute into our function:

[tex]y=6/x[/tex]

[tex]rsinu=6/rcosu[/tex]

multiply both sides by r:

[tex]r^2sinu=6/cosu[/tex]

solve for r by dividing by sinu:

[tex]r=\sqrt{6/sinucosu}[/tex]

Answer:

[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]

Step-by-step explanation:

The equation to the line normal to the curve has the following format:

[tex]y - y(x_{0}) = m(x - x_{0})[/tex]

In whicm m is the derivative of y at the point [tex]x_{0}[/tex]

In this problem, we have that:

[tex]x_{0} = \pi[/tex]

[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]

[tex]y(\pi) = 3\cos{\frac{\pi}{3}} = \frac{3}{2}[/tex]

The derivative of [tex]\cos{ax}[/tex] is [tex]a\sin{ax}[/tex]

So

[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]

[tex]y'(x) = 3*\frac{1}{3}\sin{\frac{x}{3}} = \sin{\frac{x}{3}}[/tex]

[tex]m = \sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{3}[/tex]

The equation of the line normal to the curve of y=3cos1/3x is:

[tex]y - y(x_{0}) = m(x - x_{0})[/tex]

[tex]y - \frac{3}{2} = \frac{\sqrt{2}}{3}(x - \pi)[/tex]

[tex]y = \frac{\sqrt{2}}{3}(x - \pi) +  \frac{3}{2}[/tex]

[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]

Answer:72

Step-by-step explanation:

Answer:

A researcher studies length of time in college, first through fourth year, and its relation to academic motivation. To get the most detail out of her measures, she assesses each student in both the fall and spring semesters of each of their four years in school. She finds that students have increasingly higher motivation from their first semester to their seventh semester (the start of their fourth year), with a trailing off in the last semester. The independent variable are:

Step-by-step explanation:

Answer:

If a 1% level of significance is used to test a null hypothesis, there is a probability of ____ less than 1%______ of rejecting the null hypothesis when it is true

Step-by-step explanation:

Given that a hypothesis testing is done.

Level of significance used is 1%

i.e. alpha = 1%

When we do hypothesis test, we find out test statistic Z or t suitable for the test and find p value

If p value is < 1% we reject null hypothesis otherwise we accept null hypothesis.

So p value can be atmost 1% only for accepting null hypothesis.

So the answer is 1%

If a 1% level of significance is used to test a null hypothesis, there is a probability of ____less than 1%______ of rejecting the null hypothesis when it is true.

Answer:

Step-by-step explanation:

i) Confidence level = 90%

ii) therefore significance level = [tex]\alpha[/tex]  =  [tex]\frac{100 - 90}{100}\hspace{0.1cm} = \hspace{0.1cm} 0.1[/tex]

iii) this problem is for a two tailed test.

iv) therefore significance level = [tex]\frac{\alpha }{2}[/tex] = [tex]\frac{0.1}{2}[/tex] = 0.05

iii) sample size = 15

iv) therefore degrees of freedom  = sample size  - 1   =   15 - 1    =  14

v) the upper ( or right) chi squared value, [tex]\chi^2_{R}[/tex]  from Chi squared critical value tables = 23.685

The correct critical value χR is 23.685.

To find the critical value χR corresponding to a sample size of 15 and a confidence level of 90%, we first determine the degrees of freedom.

The degrees of freedom (df) for a chi-square distribution is calculated as n - 1, where n is the sample size. Here, n = 15, so df = 14.

Next, look up the critical value for the chi-square distribution with 14 degrees of freedom and a 90% confidence level.

In statistical tables, the value corresponding to a 90% confidence level (or equivalently, 10% significance level, which leaves 5% in each tail of the distribution) for 14 degrees of freedom is 23.685.

Therefore, the correct critical value χR is 23.685.

To determine the volume of water evaporated from the pool at the El Cheapo Motel, we converted all measurements to a common unit and calculated the volume of water evaporated. The motel has to add approximately 946.13 gallons of water weekly, considering there is no rain.

To answer this question, we first need to convert the measurements to a common unit. Given that the pool is 5.0 m wide and 12.0 m long (a total area of 60.0 m2) and the weekly evaporation is 2.35 inches, we first convert the inches to meters. Since 1 inch is equal to 0.0254 meters, 2.35 inches equals 0.05969 meters.

Then, we calculate the volume of water evaporated in a week, which is calculated by multiplying the surface area of the pool by the depth of the water evaporated. Hence, it's 60.0 m2 * 0.05969 m = 3.58 m3. As 1 m3 is approximately 264.17 gallons, 3.58 m3 equals 946.1296 gallons approximately.

In conclusion, the El Cheapo Motel needs to add around 946.13 gallons of water to their pool on a weekly basis, if there is no rain.

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Answer:

Step by step approach is as shown

Step-by-step explanation:

recalling from commutative law that A + B = B + A and since A is a scalar, and from scalar multiplication of matrix.

Answer:

its 5000

Step-by-step explanation:

for it to be 50000 like the other person said it would be asking what is 0.5 not 0.05 like it says in the question

Answer:

number of subsets of a set with 13 elements are: [tex]2^{13}[/tex]

Step-by-step explanation:

In order to solve this intuitively, we can start by a set with lesser elements. This will reveal a pattern that will be used to solve for the subsets of the 13 element set.

If we start with a set B. which contains only 3 elements.

[tex]B = \{1,2,3\}[/tex]

how many subsets of B are there? well we can count them. [the set containing {1,2} and {2,1} are the same, arrangement doesn't matter]

[tex]B_{0} = \{\}\\B_{1a}=\{1\}\\B_{1b}=\{2\}\\B_{1c}=\{3\}\\B_{2a}=\{1,2\}\\B_{2b}=\{2,3\}\\B_{2c}=\{3,1\}\\B_{3a}=\{1,2,3\}\\[/tex]

there are a total of 9 subsets here.

Similarly, if you try a with a subset with only two elements you'll find that it has a total of 4 subsets.

We can see that combinatorics is at play here.

for the set B. the number of subsets can be written as:

[tex]\text{\# of subsets of B} = ^3C_0+^3C_1+^3C_2+^3C_3\\\text{\# of subsets of B} = 1+3+3+1\\\\text{\# of subsets of B} = 8[/tex]

if we try with a 2-element set:

[tex]\text{\# of subsets} = ^2C_0+^2C_1+^2C_2\\\text{\# of subsets} = 1+2+1\\\ \text{\# of subsets} = 4[/tex]

We can use the same technique to find the number of subsets of the 13 element set.

But if you recognize a pattern here that this sets of combinations are actually part of the pascal triangle, the sum of each row of the triangle is 2^{the row's number}. hence.

[tex]\text{\# of subsets of B} = 2^3\\\ \text{\# of subsets of B} = 8[/tex]

So finally, the subsets of a 13-element set A will be

[tex]\text{\# of subsets of A} = ^{13}C_0+^{13}C_1+^{13}C_2+^{13}C_3\cdots+^{13}C_{12}+^{13}C_{13}\\OR\\\text{\# of subsets of A} = 2^{13}\\\text{\# of subsets of A} = 8192[/tex]

If the set A has 13 elements, the number of different subsets is [tex]2^{13}=8192[/tex]

All the possible subsets that can be formed from any given set is called the Power set of that set. Generally, if we had a set [tex]H[/tex] such that

[tex]|H|=k[/tex]

Where [tex]|H|[/tex] denotes the cardinality, or number of elements, in [tex]H[/tex], the power set of [tex]H[/tex], denoted by [tex]P(H)[/tex], has the following formula

[tex]P(H)=2^k\text{ elements}[/tex]

So, given the set [tex]A[/tex] such that

[tex]|A|=13[/tex]

the power set of [tex]A[/tex] will have [tex]2^{13} \text{ or } 8192 \text{ elements}[/tex]

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Answer:

a) 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

b) 95.44% probability that the mean height x is between 68 and 70 inches.

Step-by-step explanation:

To solve this question, it is important to know the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that:

[tex]\mu = 69, \sigma = 2[/tex]

(a) What is the probability that an 18-year-old man selected at random is between 68 and 70 inches tall?

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68. So

X = 70

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 69}{2}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915.

X = 68

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{68 - 69}{2}[/tex]

[tex]Z = -0.5[/tex]

[tex]Z = -0.5[/tex] has a pvalue of 0.3085.

So there is a 0.6915 - 0.3085 = 0.383 = 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

(b) If a random sample of sixteen 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches?

Now we use the Central Limit Theorem, with [tex]n = 16, s = \frac{2}{\sqrt{16}} = 0.5[/tex]

The probability is also the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68, but with s as the standard deviation. So

X = 70

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{70 - 69}{0.5}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

X = 68

[tex]Z = \frac{X - \mu}{0.5}[/tex]

[tex]Z = \frac{68 - 69}{0.5}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

So there is a 0.9772 - 0.0228 = 0.9544 = 95.44% probability that the mean height x is between 68 and 70 inches.

In the case of normal distribution, the probability that a random 18-year-old man's height is between 68 and 70 inches is 0.383, while the probability that the mean height of a random sample of 16 men falls in the same range is approximately 0.999.

This question pertains to the statistical concept of normal distribution.

(a) To find the probability that a randomly selected 18-year-old man is between 68 and 70 inches tall, we first need to convert these heights to Z-scores. Given the mean height is 69 inches and the standard deviation is 2 inches, the Z-score for 68 inches is [tex](68-69)/2=-0.5[/tex] and for 70 inches it is [tex](70-69)/2=0.5[/tex]. The area between these Z-scores on a standard normal distribution chart represents the probability of a man being between 68 and 70 inches tall. This probability is approximately 0.383.

(b) In this part, we are dealing with a sample mean rather than individual values. Because of the Central Limit Theorem, a distribution of sample means will have a standard deviation equal to the population standard deviation divided by the square root of the sample size. So, our new standard deviation becomes 2/√16 = 0.5. Then we compute the Z-scores for 68 inches and 70 inches with this new standard deviation, and find the probability corresponding to the area between these Z-scores, which is considerably higher than individual case, almost 0.999.

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Answer:

A.positive and fairly strong

Step-by-step explanation:

Since an increase in food quantity usually means an increase in price, quantity and price are directly proportional, which configures a positive correlation.

Since it is stated that this relationship in observed at most restaurants, it can be concluded that there is a fairly strong correlation between  the amount of food served per person and the price paid for it.

Therefore, the answer is, A.positive and fairly strong

Answer:

The answer is A which is Positive and fairly strong.

Step-by-step explanation:

To properly do justice to the selected answer above, we describe a simple scenario.

Imagine yourself, content and perhaps a bit over-full after a lovely meal at a local restaurant. Then, the extravagant bill arrives.

Does  this high cost affirm your belief that this meal was valuable and thereby  influence your reordering of it?

Or does the cost of the meal overshadow your  enjoyment of it and leave you wishing you had chosen a simple meal at a better price point?

What features must an expensive restaurant  provide you over a bargain one to justify the extra cost?

Based of the question asked in respect to the scenario above, researchers went on a an experiment collecting different data from a restaurant and used the Latent Dirichlet Allocation (LDA) model to classify each review of the data and key words like ( e.g. food, service, price, ambiance, anecdotes, miscellaneous) were paid attention to.

After analysis, it was perceived that the value is such a tricky parameter to measure, marketers,  restaurateurs and economists often overlook it, instead focusing on objective  restaurant price and quality’s effect on customer satisfaction.

To solve for X, you first add the integers together and the X variables together. Then you subtract 3 from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 154

3X + 3 = 154

3X + 3 - 3 = 154 - 3

3X = 151

3X/3 = 151/3

X = 50 1/3

Since 50 1/3 is not an integer, there is no true answer to this problem.

To solve for three consecutive even integers where the sum of the first two equals 154, we assign x, x + 2, and x + 4 to those integers. Solving the equation results in x = 76, thus the three integers are 76, 78, and 80.

In Mathematics, specifically in algebra, this problem involves solving for unknowns. Let's call our three consecutive even numbers x, x + 2, and x + 4. This is because two integers are even if they differ by 2. The equation given by the problem is x + (x + 2) = 154, as we are told that the sum of the first and second of the three numbers is 154.

To solve this equation, firstly, combine like terms, you get 2x + 2 = 154. Secondly, isolate the variable x by subtracting 2 from both sides of this equation, you get 2x = 152. Thirdly, to solve for x, divide both sides of the equation by 2. The result is x = 76. This gives you the first integer. Since we know that the integers are consecutive even numbers, the other two integers are 76 + 2 = 78 and 76 + 4 = 80. Therefore, the three consecutive even integers are 76, 78, and 80.

Learn more about Consecutive Even Integers here:

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Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution [tex]B(n,p)[/tex], where [tex]n[/tex] is the number of trials and [tex]p[/tex] is the probability of success. It can be approximated by the Normal distribution with the mean of [tex]np[/tex] and the variance of [tex]np(1-p)[/tex], denoted by [tex]N(np,np(1-p))[/tex] if the following condition is met:

[tex]n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)[/tex]

2. Poisson distribution: Let the sampling distribution be the Poisson distribution [tex]P(\lambda)[/tex] where [tex]\lambda[/tex] is its mean. It can be approximated by the Normal distribution with the mean [tex]\lambda[/tex] and the variance [tex]\lambda[/tex], denoted by [tex]N(\lambda,\lambda)[/tex] when [tex]\lambda[/tex] is large enough, say [tex]\lambda>1000[/tex] (however, different sources may give different lower value for [tex]\lambda[/tex] but the greater it is, the better the approximation).

Answer:

[tex]182.167-2.03\frac{114.05}{\sqrt{36}}=143.580[/tex]  

[tex]182.167+2.03\frac{114.05}{\sqrt{36}}=220.754[/tex]  

So on this case the 95% confidence interval would be given by (143.580;220.754)  

Step-by-step explanation:

Assuming the following dataset:

77, 349,417,349, 167 , 225, 265, 360,205

145,335,40,139, 177,108, 163, 202, 22

123,439, 125,135, 86,43, 217,49, 156

119,178, 151, 61, 350, 312, 91, 89,89

We can calculate the sample mean with the followinf formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}= 182.167[/tex]

And the sample deviation with:

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}=114.05[/tex]

The sample size on this case is n =36.

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=182.167[/tex] represent the sample mean  

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

s=114.05 represent the sample standard deviation  

n=36 represent the sample size    

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 point estimate of the population mean is [tex]\hat \mu = \bar X =182.167[/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=36-1=35[/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(0.025,35)".And we see that [tex]t_{\alpha/2}=2.03[/tex]  

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

[tex]182.167-2.03\frac{114.05}{\sqrt{36}}=143.580[/tex]  

[tex]182.167+2.03\frac{114.05}{\sqrt{36}}=220.754[/tex]  

So on this case the 95% confidence interval would be given by (143.580;220.754)  

The question is asking for a 95% confidence interval for the mean weight of bears in a park. The confidence interval is a range of values, derived from the data collected, that is estimated to contain the true population mean. 95% of such confidence intervals are expected to contain the true value.

In statistics, we often

use sample data to make generalizations

about an unknown population. This part of statistics is known as

inferential statistics

. The sample data help us to make an estimate of a population parameter. We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, which are often called confidence intervals.

A confidence interval is a type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter.

In this case, you've been asked to calculate a 95% confidence interval for the mean weight of bears in a park. From the provided data, you would calculate the sample mean and standard deviation, and use a statistical formula to calculate the confidence interval. For example, if the confidence level is 95 percent, then we say, 'We estimate with 95 percent confidence that the true value of the population mean is between x and y.'.

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Step-by-step explanation:

The largest whole number can be 9,999,999,999 if it is  decimal representation is simple. The device with me  supports up to 10 ^ 100 exponents, so that 9.9999999E99 could be a candidate too. If your exponents are not limited, then 9E9999999 is the largest (above what the calculator can demonstrate).

Answer:

Step-by-step explanation:

The relative frequencies are 0.15, 0.08, 0.39, and 0.38 for the respective classes while 38% of the observations are at least 300 but less than 350.

The cumulative relative frequency distribution given can be used to construct a relative frequency distribution. To find the relative frequency for each class, subtract the cumulative frequency of the previous class from the cumulative frequency of the current class. Therefore:

The Total relative frequency is the sum of all the relative frequencies and should equal 1 (or almost equal to 1 due to rounding differences)

For the second part of your question: 'what percent of the observations are at least 300 but less than 350?', since this is the relative frequency distribution, it means that class 300 to 350 has already been expressed as a percentage (since relative frequency is a proportion and can be expressed as a percentage). So, 38% of the observations are at least 300 but less than 350.

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Answer:

see the picture of work shown

Answer:she has 5 dimes and 15 quarters.

Step-by-step explanation:

A dime is worth 10 cents. Converting to dollars, it becomes

10/100 = $0.1

A quarter is worth 25 cents. Converting to dollars, it becomes

25/100 = $0.25

Let x represent the number of dimes that she has.

Let y represent the the number of quarters that she has

Patrice Patriot has a total of 20 coins. It means that

x + y = 20

The total worth of dimes and quarters that she has in a piggy bank is $4.25. It means that

0.1x + 0.25y = 4.25 - - - - - - - - - -1

Substituting x = 20 - y into equation 1, it becomes

0.1(20 - y) + 0.25y = 4.25

2 - 0.1y + 0.25y = 4.25

- 0.1y + 0.25y = 4.25 - 2

0.15y = 2.25

y = 2.25/0.15

y = 15

x = 20 - y = 20 - 15

x = 5

Answer:

[tex]lim_{n \to \infty} A_n = \frac{R}{i}[/tex]

Step-by-step explanation:

For this case we have this expression:

[tex] A_n = R [\frac{1 -(1+i)^{-n}}{i}][/tex]

The lump sum investment of An is needed to result in n periodic payments of R when the interest rate per period is i.

And we want to find the:

[tex] lim_{n \to \infty} A_n[/tex]

So we have this:

[tex] lim_{n \to \infty} A_n = lim_{n \to \infty}R [\frac{1 -(1+i)^{-n}}{i}] [/tex]

Then we can do this:

[tex] lim_{n \to \infty} A_n = lim_{n \to \infty} R [\frac{1 -\frac{1}{(1+i)^n}}{i}][/tex]

[tex]lim_{n \to \infty} A_n = R lim_{n \to \infty} [\frac{1 -\frac{1}{(1+i)^n}}{i}][/tex]

And after find the limit we got:

[tex] lim_{n \to \infty} A_n = R [\frac{1-0}{i}][/tex]

Becuase : [tex] \frac{1}{(1+i)^{\infty}} =0[/tex]

And then finally we have this:

[tex]lim_{n \to \infty} A_n = \frac{R}{i}[/tex]