A random sample of 30 varieties of cereal was selected. The average number of calories per serving for these cereals is x-120. Assuming that σ = 10, find a 95% confidence interval for the mean number of calories, μ, in a serving of cereal. a. 115.30 to 124.70 b. 116.42 to 123.58 c. 118.00 to 122.00 d. 117.00 to 123.00

Answer:

-0.4

B. The answer is the same irrespective of the sample sizes.

Step-by-step explanation:

Given that an article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as 4.1 hours and 4.5 hours, respectively

Let X be the sample average lifetime of 81 Duracell and Y be the sample average lifetime of 81 Eveready Energizer batteries.

Now we have sample mean difference does not depend about the sample sizes.

This is because

E(X-Y) = E(X)-E(Y) for all X and Y

Mean of X-Y = [tex]4.1-4.5=-0.4[/tex]

This does not depend on the specified sample sizes

B. The answer is the same irrespective of the sample sizes.

The mean value of X - Y is 0.4 hours. The answer does not depend on the sample sizes.

The mean value of X - Y, where X is the sample average lifetime of 81 Duracell batteries and Y is the sample average lifetime of 81 Eveready Energizer batteries, is given by the difference in the population average lifetimes of the two types of batteries. In this case, the difference is 4.5 - 4.1 = 0.4 hours. So the distribution of X - Y is centered around 0.4 hours.

The answer does not depend on the specified sample sizes in this case. The mean value of X - Y remains the same regardless of the sample sizes, as long as the samples are representative of the populations and the population parameters do not change. So the answer is B. The answer is the same irrespective of the sample sizes.

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Answer:

Reject the null hypothesis. There is sufficient evidence to prove that the mean life is different from 7463 hours.

95% confidence interval also supports this result.

Step-by-step explanation:

Let mu be the population mean life of a large shipment of CFLs.

The hypotheses are:

[tex]H_{0}[/tex]: mu=7463 hours

[tex]H_{a}[/tex]: mu≠7463 hours

Test statistic can be calculated using the equation:

z=[tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] where

Then z=[tex]\frac{7163-7463}{\frac{1080}{\sqrt{81} } }[/tex] = -2.5

p-value is  0.0124, critical values at 0.05 significance are ±1.96

At the 0.05 level of​ significance, the the result is significant because 0.0124<0.05. There is significant evidence that mean life of light bulbs is different than 7463 hours.

95% Confidence Interval can be calculated using M±ME where

margin of error (ME) from the mean can be calculated using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

Then ME=[tex]\frac{1.96*1080}{\sqrt{81} }[/tex] =235.2

Thus 95% confidence interval estimate of the population mean life of the light bulbs is 7163±235.2 hours. That is between 6927.8 and 7398.2 hours.

Answer:

C. There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Step-by-step explanation:

1) Data given and notation  

n=1016 represent the random sample taken    

X=535 represent the people stated that they were worried about having enough money to live comfortably in retirement

[tex]\hat p=\frac{535}{1016}=0.527[/tex] estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement

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

Confidence =0.99 or 99%

z would represent the statistic

p= population proportion of people stated that they were worried about having enough money to live comfortably in retirement

2) Confidence interval

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 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]0.527 - 2.58 \sqrt{\frac{0.527(1-0.527)}{1016}}=0.487[/tex]

[tex]0.527 + 2.58 \sqrt{\frac{0.527(1-0.527)}{1016}}=0.567[/tex]

And the 99% confidence interval would be given (0.487;0.567).

There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

To build a 99% confidence interval, we first calculate our sample proportion by dividing the number of such instances by the total sample size. Next, we determine the standard error of the proportion, then our margin of error by multiplying the standard error by the Z value of the selected confidence level. Lastly, we determine the confidence interval by adding and subtracting the margin of error from the sample proportion.

To construct a 99% confidence interval for the proportion of adults worried about having enough money to live comfortably in retirement, we will utilize statistical methods and proportions. First, we must calculate the sample proportion. The sample proportion (p) is equal to 535 (the number who are worried) divided by 1016 (the total number of adults surveyed).

Then, we find the standard error of the proportion which we get by multiplying the square root of ((p*(1-p))/n) where n is the number of adults sampled. The margin of error is found using the Z value corresponding to the desired confidence level, in this case, 99%. Multiply the standard error by this Z value. Lastly, we construct the confidence interval by taking the sample proportion (p) ± the margin of error.

The result will give you the 99% confidence interval - meaning we are 99% confident that the true proportion of adults who are worried about having enough money to live comfortably in retirement lies within this interval.

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Answer:

d. 99%

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]p = 0.30, n = 50[/tex]

Let's start from the higher confidence levels, since the higher the confidence level, the higher the width of the interval.

d. 99%

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.905[/tex], so [tex]Z = 2.575[/tex].

The lower limit is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.30 - 2.575\sqrt{\frac{0.3*0.7}{50}} = 0.133[/tex]

The upper limit is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.30 + 2.575\sqrt{\frac{0.3*0.7}{50}} = 0.467[/tex]

This is very close to the interval found, so d. is the correct answer

Answer:

x + 3y - 16 = 0

Step-by-step explanation:

When two points, say [tex]$ (x_1, y_1) $[/tex] and [tex]$ (x_2, y_2) $[/tex] are given, the equation is determined using Two - point form.

The two - point form is as follows:

[tex]$ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} $[/tex]

Here: [tex]$ (x_1, y_1) = (-6, 7) $[/tex] and [tex]$(x_2, y_2) = (-3, 6) $[/tex].

Substituting in the formula we get the equation of the line as:

[tex]$ \frac{y - 7}{6 - 7} = \frac{x + 6}{- 3 + 6}[/tex]

[tex]$ \implies \frac{y - 7}{- 1} = \frac{x + 6}{ 3} $[/tex]

[tex]$ \implies 3y - 21 = - x - 6 $[/tex]

Rearranging we get: x + 3y - 15 = 0

This is the required equation of the line.

Answer:

The necessary sample size should be at least 423.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the length of the sample.

In this problem, we have that:

[tex]M = 2, \sigma = 25[/tex]. So

[tex]2 = 1.645*\frac{25}{\sqrt{n}}[/tex]

[tex]2\sqrt{n} = 41.125[/tex]

[tex]\sqrt{n} = 20.5625[/tex]

[tex]\sqrt{n}^{2} = (20.5625)^{2}[/tex]

[tex]n = 422.81[/tex]

The necessary sample size should be at least 423.

Answer:

Maximum profit is $87 when 3 blenders and 11 mixers are produced.

Step-by-step explanation:

let blender is represented by [tex]x_{1}[/tex] and and mixer by [tex]x_{2}[/tex].

total time to deliver parts = 24 hrs

total time to assemble = 30 hrs

time taken by each blender to deliver parts = 1 hr

time taken by each mixer to deliver parts = 2 hr

time taken by blenders in final assembling= 2 hr

time taken by mixers in final assembling = 3 hr

Each blender produced nets the firm=  $7

Each mixer produced nets the firm=  $6

Using this all data linear system of equation will be:

[tex]x_{1} + 2x_{2} =24  ----- (1)\\2x_{1} + 2x_{2} = 30 ----- (2)\\[/tex]

profit function:

[tex]z= 7x_{1} +6x_{2} --- (3)[/tex]

[tex]from (1)\\x_{1} = 0 \implies x_{2}= 12\\x_{2}= 0 \implies x_{1}= 24\\[/tex]

Coordinate points obtained from (1) are (0,12) and (24,0)

[tex]from (2)\\x_{1}=0 \implies x_{2}=10\\x_{2}=0 \implies x_{1}=15\\[/tex]

Coordinate points obtained from (2) are (0,10) and (15,0)

plotting these on graph

points lying in feasible region are:

A(0,0)

B(0,10)

C(3,11)

D(12,0)

substituting these points in (3) to find the maximum profit:

for A (0,0)

z = 0

for B (0,10)

z = 60

for C (3,11)

z =  87

for D (12,0)

z=84

So maximum profit is $87 when 3 blenders and 11 mixers are produced.

Answer:

If (a and b )≤ 0 then [tex]log(ab)=log(a)+log(b)[/tex] is disproved

Step-by-step explanation:

If a and b are positive real numbers then:

[tex]log(ab)=log(a)+log(b)[/tex]

But if a and b are negative then this axiom is not true as log is not defined

[tex]log_{c}(x)[/tex]= undefined   [tex]for \quad x\leq 0[/tex]

So if (a and b )≤ 0 then [tex]log_{c}(a)[/tex] and  [tex]log_{c}(b)[/tex] are undefined but [tex]log_{c}(-a*-b)[/tex] is defined.

The statement 'If a and b are any two real numbers, then log(ab) = log(a) + log(b)' is a basic property of logarithms and it is true. It represents the concept of 'logarithms of products' and falls under the wider subject of exponentials and logarithms.

The provided statement 'If a and b are any two real numbers, then log(ab) = log(a) + log(b)' is actually a basic property of logarithms. If a and b are any two positive real numbers, then the logarithm of the product of these two numbers (ab) is indeed equal to the sum of the logarithm of the first number (a) and the logarithm of the second number (b). This is mathematically represented as log(ab) = log(a) + log(b).

This property of logarithms comes under the concept of 'logarithms of products', which is a part of the wider topic of exponentials and logarithms. Using similar properties, we can say that the logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers. Also, the logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number.

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Answer:

b0= 144.59

b= -2.12

Se²= 1.02

99%CI E(Y/X=35): [68.78; 71.99]

Step-by-step explanation:

Hello!

I've arranged the given data:

X: 37.0, 36.4, 35.8, 34.3, 33.7, 32.1, 31.5

Y: 65.0, 67.2, 70.3, 71.9, 73.8, 75.7, 77.9

The equation of the linear regression model is:

Yi= β₀ + βXi + εi

Where

Yi is the dependent variable

Xi is the independent variable

εi represents the errors or residues

β₀ is the intercept of the line

β is the slope

The conditions to make a linear regression analysis are:

For each given value of X, there is a population of Y~N(μy;σy²)

Each value of Y is independent of the others.

The population variances of each population of Y are equal.

From these conditions the following characteristic is deduced:

εi~N(0;σ²)

The parameters of the regression are:

β₀, β, and σ²

If the conditions are met then you can estimate the regression line:

Yi= bo * bXi + ei.

And the point estimation of the parameters can be calculated using the formulas:

β₀ ⇒ b0= (∑y/n)-b(∑x/n)

β ⇒ b= [∑xy- ((∑x)(∑y))/n]/(∑x²-((∑x)²/n))

σ²⇒ Se²= 1/(n-2)*[∑y²-(∑y)²/n - b²(∑x²-(∑x)²/n)]

n= 7

∑y= 501.80

∑y²= 36097.88

∑x= 240.80

∑x²= 8310.44

∑xy= 17204.87

b0= 144.59

b= -2.12

Se²= 1.02

The estimated regression line is:

Yi= 144.59 -2.12Xi

You need to calculate a 99%CI E(Y/X=35), the formula is:

(b0 + bX0) ± [tex]t_{n-2;1-\alpha /2}[/tex]*[tex]\sqrt{S_e^2(\frac{1}{n}+\frac{(X_0-X[bar])^2}{sumX^2-(\frac{(sumX)^2}{n} )} )}[/tex]

(144.59 + (-2.12*35)) ± 4.032*[tex]\sqrt{1.02(\frac{1}{7}+\frac{(35-34.4)^2}{8310.44-(\frac{(240.80)^2}{7} )} )}[/tex]

[68.78; 71.99]

With a 99% confidence level youd expect that the interval [68.78; 71.99] contains the true value of the average of Y when X= 35.

I hope it helps!

Answer:

19

Step-by-step explanation:

Given:

measuring cup size = 1/4 cup

total amount to be measured,

= 4 3/4 cups (convert to improper fraction)

= 19/4 cups.

number of 1/4 cups

= total amount measured ÷ 1/4 cup

= 19/4 ÷ 1/4

= 19/4 x 4/1

= 19/4 x 4

= 19         (1/4-cups) used

To figure out how many 1/4 cups Mel will need, we need to convert 4 3/4 cups to quarters. As 4 cups equal 16 quarters and 3/4 equals 3 quarters, we have a sum of 19 quarters. Therefore, Mel will need 19 measures of his 1/4 cup.

To solve this problem, you need to understand how to convert measurements using fractions. We know that Mel needs 4 3/4 cups of flour and the only measuring cup he has measures 1/4 cup.

The first thing we can do is convert 4 3/4 into an improper fraction. One whole cup is 4 quarters, so 4 cups are 4 x 4 = 16 quarters. Additionally, there are 3 more quarter cups, totaling 16 + 3 = 19 quarter cups.

So, Mel needs 19 quarter cups of flour.

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Candy can arrange in 1, 2, 3, 5, 6, 10, 15 ways

2, 3, 4, 6, 8, 12 number of people can sit at each table

Solution:

Candy brought 30 cookies to a party. List all the ways she can arrange them on a plate in equal rows.

Given that Candy brought 30 cookies to party

She has to arrange the ways she can arrange them on plate in equal rows

We have to list the factors of 30

factors of 30 = 1, 2, 3, 5, 6, 10, 15

So she can arrange in 8 ways

Possible arrangement are 1, 2, 3, 5, 6, 10, 15 rows

David is setting up tables for 24 people at the party. The same number of people will sit at each table, and no one will sit alone. How many people can sit at each table? List all possibilities.

Fcators of 24 = 1, 2, 3, 4, 6, 8, 12

Here we can eliminate 1 since given that no one will sit alone

Thus 2, 3, 4, 6, 8, 12 number of people can sit at each table

Answer:

[tex]0.0864 < p< 0.1358[/tex]

We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.  

Step-by-step explanation:

1) Data given and notation  

n=621 represent the random sample taken    

X=69 represent the people admitted they did not buy a ticket

[tex]\hat p=\frac{69}{621}=0.111[/tex] estimated proportion of people admitted they did not buy a ticket

[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)    

z would represent the statistic

p= population proportion of people admitted they did not buy a ticket

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".

2) Confidence interval

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]0.111 - 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.0864[/tex]

[tex]0.111 + 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.1358[/tex]

And the 95% confidence interval would be given (0.0864;0.1358).

[tex]0.0864 < p< 0.1358[/tex]

We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.  

Answer: The correct answer is C.

Step-by-step explanation: The y-intercept is the point on the graph at which the line crosses over the y-axis. In this case, your y-intercept is 1, because the line crosses over the y-axis at 1. The slope formula is rise/run, or y2 -y1/x2-x1. However since you were not supplies with any coordinates, you'll have to go with rise (up/down) over run (left/right). The slope of the line falls one cube down ward, making the slope negative 1. The slope  "runs" to the right 4 cubes, making the slope -1 (falls 1 cube)/4 (moves to the left/right 4 cubes).

Answer:

CD ≅ CD, Reflexive property

Step-by-step explanation:

We want to show the triangles are similar by SAS.  We've already proven that one pair of their sides are congruent (AC ≅ BD), and that one pair of their angles are congruent (∠CDE ≅ ∠DCE), so we need to show that the next pair of sides are congruent.

From the two column proof below, we have seen the missing step 8 is: CD ≅ CD, Reflexive property

The two column proof to show that ΔACD ≅ ΔBCD

We are given that AC ≅ BD because congruent segments added to congruent segments form congruent segments.

We are also given that ∠CDE ≅ ∠DCE. This means we have one side and the included angle as congruent.

We need one more congruent side to prove Congruency. Thus:

CD ≅ CD, Reflexive property

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The estimated slope is approximately -4.028 and the estimated y-intercept is approximately 92.919.

The estimated value of y when x=5 is approximately 72.779, and the error prediction when x=2 is approximately 5.863.

The estimated slope (b1) and the estimated y-intercept (bo) in the regression equation y = bo + b1x, you can use the following formulas:

[tex]\[b1 = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\][/tex]

[tex]\[bo = \frac{\sum y - b1(\sum x)}{n}\][/tex]

Where:

n is the number of data points,

[tex]\(\sum xy\)[/tex] is the sum of the product of x and y,

[tex]\(\sum x\)[/tex] is the sum of x,

[tex]\(\sum y\)[/tex] is the sum of y,

[tex]\(\sum x^2\)[/tex] is the sum of the squared x values.

Now let's calculate these values step by step:

a. Find the estimated slope (b1):

[tex]\[n = 10\][/tex]

[tex]\[\sum x = 31\][/tex]

[tex]\[\sum y = 804\][/tex]

[tex]\[\sum xy = 2439\][/tex]

[tex]\[\sum x^2 = 121.25\][/tex]

[tex]\[b1 = \frac{(10 \times 2439) - (31 \times 804)}{(10 \times 121.25) - 31^2} \][/tex]

[tex]\[b1 = \frac{24390 - 25404}{1212.5 - 961}\][/tex]

[tex]\[b1 = \frac{-1014}{251.5}\][/tex]

[tex]\[b1 \approx -4.028 \ (rounded to three decimal places)\][/tex]

b. Find the estimated y-intercept (bo):

[tex]\[bo = \frac{804 - (-4.028 \times 31)}{10}\][/tex]

[tex]\[bo = \frac{804 + 125.188}{10}\][/tex]

[tex]\[bo \approx \frac{929.188}{10}\][/tex]

[tex]\[bo \approx 92.919 \ (rounded to three decimal places)\][/tex]

c. Find the estimated value of y when x=5:

[tex]\[y = bo + b1x\][/tex]

[tex]\[y = 92.919 + (-4.028 \times 5)\][/tex]

[tex]\[y = 92.919 - 20.14\][/tex]

[tex]\[y \approx 72.779 \ (rounded to three decimal places)\][/tex]

d. Find the error prediction when x=2:

First, find the predicted y using the regression line:

[tex]\[y = bo + b1x\][/tex]

[tex]\[y = 92.919 + (-4.028 \times 2)\][/tex]

[tex]\[y = 92.919 - 8.056\][/tex]

[tex]\[y \approx 84.863 \ (rounded to three decimal places)\][/tex]

Now, find the error prediction:

[tex]\[Error = |Observed\, y - Predicted\, y|\][/tex]

[tex]\[Error = |79 - 84.863|\][/tex]

[tex]\[Error \approx 5.863 \ (rounded to three decimal places)\][/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.  

Answer:

(A) It is less than the contribution predicted by the regression line.

Step-by-step explanation:

The residual is the subtraction of the observed value by the predicted value.

So, if James has a negative residual, it means that his contribution is less than what was expected by the regression line.

The correct answer is:

(A) It is less than the contribution predicted by the regression line.

James' actual contribution must be less than the contribution predicted by the regression line.

To determine which statement must be true about James' actual contribution, we need to understand the concept of residuals in regression analysis. A residual is the difference between the actual observed value and the predicted value. Since James' residual is negative, it means that his actual contribution is less than the contribution predicted by the regression line. (A) It is less than the contribution predicted by the regression line.

However, we cannot determine if James' actual contribution is less than the average contribution made by the 50 alumni solely based on the fact that his residual is negative. The regression line only predicts individual contributions based on the number of years since graduation, and the average contribution is not directly related to the regression line. Therefore, (B) It is less than the average contribution made by the 50 alumni. cannot be concluded.

Therefore, the correct answer is (A) It is less than the contribution predicted by the regression line.

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The expected population of the Midwest city in 2036, based on the given parameters of an initial population (P₀) of 307,000 people in 2014, a yearly growth rate (r) of 0.673%, and a time period (t) of 22 years, is approximately 355,992.07.

Find the expected population of the Midwest city in 2036:

1. Define the variables:

P: Population in the year 2036 (unknown)

P₀: Initial population in 2014 (307,000 people)

t: Number of years since 2014 (2036 - 2014 = 22 years)

r: Yearly growth rate (0.673% = 0.00673 as a decimal)

2. Apply the formula:

The formula for exponential population growth is:

P = P₀ * e^(r * t)

where:

e is the base of the natural logarithm (approximately 2.71828)

3. Plug in the values:

P = 307,000 * e^(0.00673 * 22)

4. Calculate the result:

Using a calculator or spreadsheet, we get:

P ≈ 355,992.07

Answer:

Probability=1-[tex]\frac{(5!)^{3} }{15!}[/tex]

Step-by-step explanation:

We have to solve this question with the help of complementary method. Observe the statement "Probability that every family has at least two of its children in the same tent". The complementary of this statement will be every family does not have atleast two of its children in the same tent ie. Each child of a family is not in the same tent as one from the same family.

P(A)=1-[tex]P(A)^{c}[/tex]

Therefore, we can get P(A) if we just take out the value of [tex]P(A)^{c}[/tex]

Probability=[tex]\frac{TotalNo.OfFavourableOutcomes}{TotalNo.ofOutcomes}[/tex]

Imagine that we have 15 locations to fill and 15 people for it, so the total no. of cases= 15!

Bernstein's children can be arranged in 5 different tents in 5! ways.

Similarly Henderson's and Smith's children can be arranged in 5 different tents in 5! ways only.

Therefore, [tex]P(A)^{c}[/tex]= [tex]\frac{(5!)^{3} }{15!}[/tex]

P(A)=1- [tex]\frac{(5!)^{3} }{15!}[/tex]

Answer:

[tex]SE=\sqrt{\frac{0.505 (1-0.505)}{832}+\frac{0.0311(1-0.0311)}{145}}=0.0226[/tex]

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".  

1) Data given and notation  

[tex]X_{1}=420[/tex] represent the number of cats diagnosing malignant lymphoma from homes where herbicides were used​ regularly

[tex]X_{2}=17[/tex] represent the number of cats diagnosing malignant lymphoma from homes where NO herbicides were used​ regularly

[tex]n_{1}=832[/tex] sample 1 selected

[tex]n_{2}=145[/tex] sample 2  selected

[tex]\hat p_{1}=\frac{420}{832}=0.505[/tex] represent the proportion of of cats diagnosing malignant lymphoma from homes where herbicides were used​ regularly

[tex]\hat p_{2}=\frac{17}{145}=0.0311[/tex] represent the proportion of cats diagnosing malignant lymphoma from homes where NO herbicides were used​ regularly

z would represent the statistic (variable of interest)  

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

[tex]p_1 -p_2[/tex] parameter of interest

2) Solution to the problem

We are interested on the standard error for the difference of proportions and is given by this formula:

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

And if we replace the values given we got:

[tex]SE=\sqrt{\frac{0.505 (1-0.505)}{832}+\frac{0.0311(1-0.0311)}{145}}=0.0226[/tex]

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-6 from both sides

3/x-2 = √(x - 2) + 2

Multiply both sides by (x - 2)

(Note: this cancels out the square root)

3 = x - 2 + 2

x = 3

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Answer: 33 bananas

Step-by-step explanation:

Since the monkey can only carry 100 bananas at a time.

And a total of 200 bananas is to be transferred through 100yards.

With the condition that he eats 1 banana per yard .

If he decides to carry 100 banana straight through the whole 100yards. He would be left with no banana at the end of the journey and will not have enough banana to be able to come back and carry the rest.

Therefore, the best way is to transfer all the banana to a destination where he will be left with 100bananas that he can then move at once to the final destination.

To move 200 bananas to a particular destination given that he can carry only 100 banana each.

He would need to travel twice that means ( three trips i.e leave 100 return with 0 leave 0)

For three trips through a particular length in yards to exhaust 100 yards. The length in yards is given as

100 = 3 × l

l = 100/3

l = 33 yards

Travelling 33 yards he would be left with 100 bananas that he can carry in just one trip through the rest of the journey which is 100 -33 yards = 67yards

Carrying 100 bananas through 67 yards in one trip

He will exhaust;

1 banana per yard × 67yards = 67 bananas

Therefore, he would be left with

100 - 67 bananas

= 33 bananas

Answer:

At least 75% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.

The minimum possible body temperature that is within 2 standard deviation of the mean is 97.02F and the maximum possible body temperature that is within 2 standard deviations of the mean is 99.54F.

Step-by-step explanation:

Chebyshev's theorem states that, for a normally distributed(bell-shaped )variable:

75% of the measures are within 2 standard deviations of the mean

89% of the measures are within 3 standard deviations of the mean.

Using​ Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2 standard deviations of the​ mean?

At least 75% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.

Range:

Mean: 98.28

Standard deviation: 0.63

Minimum = 98.28 - 2*0.63 = 97.02F

Maximum = 98.28 + 2*0.63 = 99.54F

The minimum possible body temperature that is within 2 standard deviation of the mean is 97.02F and the maximum possible body temperature that is within 2 standard deviations of the mean is 99.54F.

Answer:

n=500 represent the random sample taken    

[tex]\hat p=0.12[/tex] estimated proportion of people that chose chocolate​ pie

[tex]\hat q =1-\hat p=1-0.12=0.88[/tex] represent the people that NOT chose chocolate​ pie

E=0.05 represent the error or margin of error given by the following formula:

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

p= true population proportion of people that chose chocolate​ pie

If the confidence level is 90 ​%, what is the value of alpha ​?

[tex]\alpha=1-0.9 =0.1[/tex] and the value of [tex]\alpha/2 =0.05[/tex],

[tex]z_{\alpha/2}=-1.64[/tex] and [tex]z_{1-\alpha/2}=1.64[/tex]

[tex]ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238[/tex]

Step-by-step explanation:

Data given and notation

What values do ModifyingAbove p with caret ​, ModifyingAbove q with caret ​, ​n, E, and p​ represent?

n=500 represent the random sample taken    

X represent the people that chose chocolate​ pie

[tex]\hat p=0.12[/tex] estimated proportion of people that chose chocolate​ pie

[tex]\hat q =1-\hat p=1-0.12=0.88[/tex] represent the people that NOT chose chocolate​ pie

E=0.05 represent the error or margin of error given by the following formula:

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

z would represent the quantile of the normal standard distribution

p= true population proportion of people that chose chocolate​ pie

The confidence interval for the population proportion is given by this formula :

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

If the confidence level is 90 ​%, what is the value of alpha ​?

On this case the value for the significance would be [tex]\alpha=1-0.9 =0.1[/tex] and the value of [tex]\alpha/2 =0.05[/tex], we can find the quantiles of the normal standard distribution given by:

[tex]z_{\alpha/2}=-1.64[/tex] and [tex]z_{1-\alpha/2}=1.64[/tex]

And with the following excel codes:

"=NORM.INV(0.05,0,1)" "=NORM.INV(1-0.05,0,1)"

And we can find the margin of error like this:

[tex]ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238[/tex]

Final answer:

In the poll,
ModifyingAbove p with caret and
ModifyingAbove q with caret represent the sample proportions of respondents who chose and did not choose chocolate pie, respectively. The symbol n represents the sample size, E denotes the margin of error, and p is the population proportion. If the confidence level is 90%, the value of alpha is 0.10.

Explanation:

In the context of the provided poll scenario, the various symbols represent the following statistical terms:
ModifyingAbove p with caret (
ModifyingAbove p with caret) represents the sample proportion, which is the observed percentage in the sample that chose chocolate pie. In this case, it would be 0.12 or 12%.
ModifyingAbove q with caret (
ModifyingAbove q with caret) is the sample proportion of respondents who did not choose chocolate pie, which would be 1 - 0.12 = 0.88 or 88%.

n represents the sample size, which is the number of respondents in the poll, 500 in this scenario.

E denotes the margin of error, which is
plus or minus 5 percentage points in this case.

p represents the population proportion, which is the true percentage of all adults who would choose chocolate pie as their favorite if everyone was surveyed.

If the confidence level is 90%, the value of alpha (
alpha) is the probability that the true population parameter will not be contained in the confidence interval. For a 90% confidence level, alpha would be 1 - 0.90 = 0.10 or 10%. This is often split into two tails of the normal distribution for a two-tailed test, thus each tail would have an area of 0.05.

To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean for the goods-producing industries, a hypothesis test can be conducted. Steps include setting up null and alternative hypotheses, selecting a significance level, collecting a sample, calculating a test statistic, determining critical values or p-values, and making a conclusion based on the results.

To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean of $24.57 for the goods-producing industries, we can conduct a hypothesis test.

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Answer:The type of the problem described above is a Sinking Fund

Option B

Step-by-step explanation:

In order to understand the solution to this question we have to be familiar with these concepts  

Sinking Fund

A sinking fund is an account earning compound interest into which you make periodic deposits. Suppose that the account has an annual interest rate of (r) compounded (m) times per year, so that (i=r/m) is the interest rate per compounding period. If you make a payment of PMT at the end of each period, then the future value after (t) years, or (n = mt) periods, will be  

   FV = [tex] PMT  (〖(1+i )〗^n  -1)/i [/tex]

Where FV is the amount that would be accumulated after t years    

Payment Formula for a Sinking Fund  

Suppose that an account has an annual rate of (r) compounded (m) times per year, so that is (i=r/m) is the interest rate per compounding period. If you want to accumulate a total of FV in the account after t years, or (n = mt) periods, by making payments of PMT at the end of each period, then each payment must be

                                           PMT = [tex] FV  ( i)/(〖(1+i)〗^n  -1) [/tex]

From the question  

 Rate= r = 3/100 = 0.03

Number of times it was paid (compounded) in a year = m = 4 its value is Four cause the payment is made 4 times in one year i.e. Quarterly  

The interest rate per compounding period = I = r/m = 0.03/4 = 0.0075

Number of times it was paid (compounded) t years n = 4 x 3 = 12

The amount that ted desires to be in that account  after 3 years =FV = $25,000  

   So the investment that Ted needs to make Quarterly in order to get his desired amount is  

               = [tex]25000 × (0.0075/(〖(1+0.0075)〗^12  -1 )) [/tex]

                = $2000  

The correct option is b. Sinking Fund. Ted should invest approximately $22,857.14 each quarter to have $25,000 in 3 years with an annual interest rate of 3% compounded quarterly.

To determine how much Ted should invest each quarter into an account that pays 3% per year compounded quarterly, we can use the sinking fund formula:[tex]\[ P = \frac{FV}{\left(1 + \frac{r}{m}\right)^{n \cdot m}} \][/tex]

where:

P is the payment made each period (quarterly in this case),

FV is the future value of the investment, which is $25,000,

r is the annual interest rate (3% or 0.03),

m is the number of times interest is compounded per year (4 for quarterly),

n is the number of years (3).

First, we need to adjust the interest rate for quarterly compounding:

[tex]\[ \text{Quarterly interest rate} = \frac{r}{m} = \frac{0.03}{4} = 0.0075 \][/tex]

Next, we calculate the number of total compounding periods:

[tex]\[ n \cdot m = 3 \cdot 4 = 12 \][/tex]

Now we can plug these values into the sinking fund formula to solve for P:

[tex]\[ P = \frac{25000}{\left(1 + 0.0075\right)^{12}} \][/tex]

[tex]\[ P = \frac{25000}{\left(1.0075\right)^{12}} \][/tex]

[tex]\[ P = \frac{25000}{1.0934} \][/tex]

[tex]\[ P \approx \frac{25000}{1.0934} \][/tex]

[tex]\[ P \approx 2285.71 \][/tex]

Therefore, Ted should invest approximately $22,857.14 each quarter to have $25,000 in 3 years with an annual interest rate of 3% compounded quarterly.

Answer:

Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]

Total exercises individual need=   [tex]y=\frac{4x}{2}[/tex]      

                                                         [tex]y-\frac{4x}{2}[/tex]  

Step-by-step explanation:

No. of groups = 4

Each group has to do exercises = 2

Total no.  of exercises individual need = y

Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]

Total exercises individual need= [tex]y=2\frac{x}{4}(4)[/tex]

                                                      [tex]y=\frac{4x}{2}[/tex]      

                                                       [tex]y-\frac{4x}{2}[/tex]  

Answer:

y-2{x/4}

Step-by-step explanation:

Answer:

a) 3⁵5³.

b) 1

c) 23³

d) 41·43·53

e) 1

f) 1111

Step-by-step explanation:

The greatest common divisor of two integers is the product of their common powers of primes with greatest exponent.

For example, to find gcd of 2⁵3⁴5⁸ and 3⁶5²7⁹ we first identify the common powers of primes, these are powers of 3 and powers of 5. The greatest power of 3 that divides both integers is 3⁴ and the greatest power if 5 that divides both integers is 5², then the gcd is 3⁴5².

a) The greatest common prime powers of 3⁷5³7³ and 2²3⁵5⁹ are 3⁵ and 5³ so their gcd is 3⁵5³.

b) 11·13·17 and 2⁹3⁷5⁵7³ have no common prime powers so their gcd is 1

c) The only greatest common power of 23³ and 23⁷ is 23³, so 23³ is the gcd.

d) The numbers 41·43·53 and 41·43·53 are equal. They both divide themselves (and the greatest divisor of a positive integer is itself) then the gcd is 41·43·53

e) 3³5⁷ and 2²7² have no common prime divisors, so their gcd is 1.

f) 0 is divisible by any integer, in particular, 1111 divides 0 (1111·0=0). Then 1111 is the gcd

Greatest common divisors were calculated for each of the given pairs. Many pairs had no common factors and their GCD was 1, while others had a GCD equal to a shared prime factor or to one of the pair elements.

In the subject of mathematics, specifically number theory, the greatest common divisor (GCD) is the largest number that divides two or more numbers without a remainder. Let's determine the greatest common divisors of the given pairs:

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Answer:

So on this case the 95% confidence interval would be given by [tex]-1.4485 \leq \mu_G -\mu_B \leq -0.9515[/tex]  

Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.

1. What is the value of the lower end of the confidence interval?

b. – 1.4485

2. What is the value of the upper end of the confidence interval?

c. – 0.9515

What kind of distribution ( central limit theorem or T) and why?

We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom  of the t distribution increases we have a normal distribution.

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 =5.1[/tex] represent the sample mean 1 (girls)

[tex]\bar X_2 =6.3[/tex] represent the sample mean 2  (boys)

n1=250 represent the sample 1 size  

n2=280 represent the sample 2 size  

[tex]s_1 =1.2[/tex] sample standard deviation for sample 1

[tex]s_2 =1.7[/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 =5.1-6.3=-1.2[/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=250+280-2=528[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,528)".And we see that [tex]t_{\alpha/2}=1.964[/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{1.2^2}{250}+\frac{1.7^2}{280}}=0.127[/tex]

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

[tex]-1.2-1.96\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=-1.4485[/tex]  

[tex]-1.2+1.96\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=-0.9515[/tex]  

So on this case the 95% confidence interval would be given by [tex]-1.4485 \leq \mu_G -\mu_B \leq -0.9515[/tex]  

Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.

1. What is the value of the lower end of the confidence interval?

b. – 1.4485

2. What is the value of the upper end of the confidence interval?

c. – 0.9515

What kind of distribution ( central limit theorem or T) and why?

We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom  of the t distribution increases we have a normal distribution.

By deriving and equating to zero the revenue and profit functions, the maximum revenue is $900,000 achieved at 3000 units sold. The maximum profit is $384,000 when 2400 units are sold at $180 each. If taxed $55/unit, the maximum profit is $302,400 at 1840 units with a price of $208/unit.

To solve this, we'll first calculate the revenue function R(x) which is the product of the number of units sold and the price per unit, i.e., R(x) = x*p(x). Then, we'll find the profit function P(x), which is the difference between the revenue and the cost, i.e., P(x) = R(x) - C(x). Next, to maximize revenue and profit, we'll get the derivative of both R(x) and P(x), set them equal to zero, and solve.

For part (A), R(x) = x*(300 - x/20) = 300x - x^2/20. Its derivative is R'(x) = 300 - x/10. Setting R'(x) = 0, we get x = 3000 units, which leads to the maximum revenue of $900,000.

For part (B), P(x) = R(x) - C(x) = 300x - x^2/20 - (72000 + 60x). Its derivative, P'(x), when equal to zero gives us x = 2400 units. Thus, the maximum profit is $384,000 and the price per unit is $180.

For part (C), the new cost function very becomes C(x) = 72,000 + 115x. Setting the derivative of the new profit equation P'(x) = 0, we get x = 1840 units, which leads to the maximum profit of $302,400. The company should charge $208 per unit.

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(A) Maximum revenue: $450,000

(B) Maximum profit: $216,000 at production level of 2400 sets, price per set: $180

(C) With a $55 tax per set, maximum profit is $84,000 at production level of 2400 sets, price per set remains $180.

(A) To find the maximum revenue, we first need to maximize the revenue function:

[tex]\[ R(x) = (300 - \frac{x}{20}) \times x \]\[ R(x) = 300x - \frac{1}{20}x^2 \][/tex]

Now, let's find the critical points by taking the derivative of R(X) and setting it equal to zero:

[tex]\[ \frac{dR}{dx} = 300 - \frac{1}{10}x \]\[ \frac{1}{10}x = 300 \]\[ x = 3000 \][/tex]

So, the maximum revenue occurs at \( x = 3000 \).

Plugging \( x = 3000 \) back into the revenue function:

[tex]\[ R(3000) = (300 - \frac{3000}{20}) \times 3000 \]\[ R(3000) = (300 - 150) \times 3000 \]\[ R(3000) = 150 \times 3000 = 450000 \][/tex]

The maximum revenue is $450,000.

(B) To find the maximum profit, we need to maximize the profit function P(x) :

P(x) = R(x) - C(x)

Given C(x) = 72,000 + 60x, we have:

[tex]\[ P(x) = (300 - \frac{x}{20}) \times x - (72,000 + 60x) \]\[ P(x) = (300x - \frac{1}{20}x^2) - (72,000 + 60x) \]\[ P(x) = 300x - \frac{1}{20}x^2 - 72,000 - 60x \]\[ P(x) = -\frac{1}{20}x^2 + 240x - 72,000 \][/tex]

Now, let's find the critical points by taking the derivative of P(x)  and setting it equal to zero:

[tex]\[ \frac{dP}{dx} = -\frac{1}{10}x + 240 \][/tex]

Setting dP/dx = 0:

-1/10x + 240 = 0

1/10x = 240

x = 2400

Now, we need to check the endpoints of the interval [tex]\( 0 \leq x \leq 6000 \)[/tex] for potential maximum profit.

[tex]\[ P(0) = -\frac{1}{20}(0)^2 + 240(0) - 72,000 = -72,000 \]\[ P(6000) = -\frac{1}{20}(6000)^2 + 240(6000) - 72,000 = -132,000 \][/tex]

So, the maximum profit occurs at  x = 2400.

Plugging x = 2400  back into the profit function:

P(2400) = [tex]-\frac{1}{20}(2400)^2[/tex] + 240(2400) - 72,000  

P(2400) = -288,000 + 576,000 - 72,000

P(2400) = 216,000

The maximum profit is $216,000.

To find the price per set at the optimal production level, we use the price-demand equation:

p(2400) = 300 - 2400/20

p(2400) = 300 - 120

p(2400) = 180

So, the company should charge $180 per television set to realize the maximum profit.

(C) If the government decides to tax the company $55 for each set it produces, the cost function becomes:

C(x) = 72,000 + 60x + 55x

C(x) = 72,000 + 115x

To find the new maximum profit, we repeat the steps from part (B) with the updated cost function. We already know that the optimal production level is x = 2400, so we can directly plug this value into the new profit function.

P(x) = R(x) - C(x)

P(2400) = R(2400) - C(2400)

Plugging in x = 2400  into the revenue function:

R(2400) = (300 - 2400/20 * 2400

R(2400) = (300 - 120) * 2400

R(2400) = 180 * 2400

R(2400) = 432,000

Plugging in  x = 2400 into the updated cost function:

C(2400) = 72,000 + 115 * 2400

C(2400) = 72,000 + 276,000

C(2400) = 348,000

Now, calculate the new profit:

P(2400) = R(2400) - C(2400)

P(2400) = 432,000 - 348,000

P(2400) = 84,000

So, the company should manufacture 2400 sets each month to maximize its profit. The maximum profit is $84,000.

To find the price per set at the optimal production level with the tax:

p(2400) = 300 - 2400/20

p(2400) = 300 - 120

p(2400) = 180

The company should still charge $180 per television set to realize the maximum profit, even with the tax.

Therefore,

(A) Maximum revenue: $450,000

(B) Maximum profit: $216,000 at production level of 2400 sets, price per set: $180

(C) With a $55 tax per set, maximum profit is $84,000 at production level of 2400 sets, price per set remains $180.