I need help with 8 and 9 please!

Answer: (16x - 2)inches

Step-by-step explanation:

The shape of Jake's window is a rectangle. It means that the two opposite sides are equal. The perimeter of the rectangular window is the distance round it.

The perimeter of a rectangle is expressed as 2(length + width)

From the dimensions given

Length = height of the window

= 5x - 3 inches

Width = 3x + 2 inches

Perimeter of the window =

2(5x - 3 + 3x + 2) = 10x - 6 + 6x + 4

= 10x + 6x + 4 - 6

Perimeter of the window =

(16x - 2)inches

Answer:The perimeter of window can be calculated by the following formula;

Step-by-step explanation:

Answer:

(b) 4,445

Step-by-step explanation:

If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?

Given a=0.05, |Z(0.025)|=1.96 (check standard normal table)

So n=(Z/E)^2*p*(1-p)

=(1.96/0.015)^2*0.5*0.5

=4268.444

Take n=4269

Answer:(b) 4,445

Answer:

  B. Neither plan is correct

Step-by-step explanation:

Yahto has the right idea in that opposite sides of a rectangle are the same length. However, that is also true of a parallelogram that is not a rectangle. The condition Yahto is looking for is necessary, but not sufficient.

__

Nora's plan will discover if adjacent sides are perpendicular to each other, provided that the slopes are defined in each case. Her plan will not work in the event there are vertical sides with undefined slope. Any quadrilateral in which all adjacent pairs of sides form right angles will be a rectangle.

_____

Strictly speaking, neither plan is completely correct. Yahto can discover rectangles that Nora cannot, and Nora can determine quadrilaterals are not rectangles when Yahto would improperly classify them.

_____

Comment on showing a quadrilateral is a rectangle

My favorite plan is to show the diagonals are the same length and have the same midpoint: J+L = K+M; ║J-L║ = ║K-M║.

Both Yahto and Nora's plans make sense, but only Nora's plan will always correctly determine if a quadrilateral is a rectangle. This is because finding negative reciprocal slopes confirms perpendicular, or right, angles which are necessary for a shape to be a rectangle.

Both Yahto and Nora have good strategies, but only Nora’s plan will always lead to the correct identification of a rectangle. A rectangle is defined as a quadrilateral with four right angles. While having equal length of opposite sides is a property of rectangles, it is not exclusive to rectangles. It can also be true for other quadrilaterals, such as parallelograms and rhombuses.

On the other hand, Nora's plan relies on finding negative reciprocal vertices, which is in line with a property of rectangles – adjacent sides are perpendicular. In a coordinate plane, if two lines are perpendicular, the slopes of the lines are negative reciprocals of each other. Hence, if these conditions are met, it would confirm that the quadrilateral is a rectangle.

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Final answer:

To calculate the 99% confidence interval for the population mean commute time, we use the formula CI = x± t * (s/√n), where x is the sample mean, t is the t-value, s is the sample standard deviation, and n is the sample size. By substituting the given values into the formula, we find that the confidence interval is approximately (29.4, 36.2) and the margin of error is 1.7. This means we are 99% confident that the population mean commute time lies within the interval, and the margin of error represents the possible deviation of the sample mean from the population mean.

Explanation:

To construct a 99% confidence interval for the population mean μ, we can use the formula:

CI = x± t * (s/√n)

Where CI is the confidence interval, x is the sample mean, t is the t-value, s is the sample standard deviation, and n is the sample size. In this case, x = 32.8, s = 7.2, and n = 27.

To find the t-value, we need to determine the degrees of freedom (df), which is equal to n - 1. So, df = 27 - 1 = 26. Looking up the t-value for a 99% confidence level with 26 degrees of freedom in a t-table or using calculator software, we find it to be approximately 2.787.

Substituting these values into the formula, we get:

CI = 32.8 ± 2.787 * (7.2/√27)

Calculating this, we find that the confidence interval is approximately (29.4, 36.2).

The margin of error, also known as the error bound for a population mean (EBM), is half the width of the confidence interval. So, the margin of error in this case is (36.2 - 32.8)/2 = 1.7.

Interpreting the results, we can say that we are 99% confident that the population mean μ lies within the range of 29.4 to 36.2 minutes. Furthermore, the margin of error of 1.7 indicates how much the sample mean could vary from the population mean, taking into account the variability in the data.

Answer: No,  It does not indicates that the population proportion of consumers loyal to Chevrolet is more than 47%.

Step-by-step explanation:

Let p denotes the proportion of consumers loyal to Chevrolet.

As per given , we have

[tex]H_0: p=0.47\\\\ H_a: p>0.47[/tex]

Since the alternative hypothesis [tex](H_a)[/tex] is right-tailed so the test would be a right-tailed test.

Also , it is given that ,Chevrolet did a study of a random sample of 1006 Chevrolet owners and found that 490 said they would buy another Chevrolet.

i.e. n = 1006

x= 490

[tex]\hat{p}=\dfrac{490}{1006}=0.487[/tex]

Test statistic :

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

, where p=population proportion.

[tex]\hat{p}[/tex]= sample proportion

n= sample size.

i.e. [tex]z=\dfrac{0.487-0.47}{\sqrt{\dfrac{0.47(1-0.47)}{1006}}}=1.08[/tex]

P-value (for right-tailed test)= P(z>1.08)=1-P(z≤1.08)  [∵P(Z>z)=1-P(Z≤z)]

=1- 0.8599=0.1401    [By z-value table.]

Decision : Since p-value (0.14) is greater than the significance level ([tex]\alpha=0.01[/tex]) , it means we are failed to reject the null hypothesis.

Conclusion : We have sufficient evidence to support the claim that about 47% of the general consumer population in the United States is loyal to the automobile manufacturer of their choice.

Hence, it does not indicates that the population proportion of consumers loyal to Chevrolet is more than 47%.

Final answer:

There's sufficient evidence to support the claim that about 47% of the general consumer population in the United States is loyal to their chosen automobile manufacturer. Thus, it doesn't indicate that the population proportion of consumers loyal to Chevrolet is more than 47%.

Explanation:

Given:

Null hypothesis (H0): The proportion of consumers loyal to Chevrolet is 47%.

Alternative hypothesis (Ha): The proportion of consumers loyal to Chevrolet is greater than 47%.

Data:

Chevrolet conducted a study of a random sample of 1006 Chevrolet owners.

Among them, 490 said they would buy another Chevrolet.

Calculations:

Sample proportion [tex](\hat{p}) = 490/1006 ≈ 0.487[/tex]

Test statistic (z) =  [tex](0.487 - 0.47) / \sqrt((0.47 * (1 - 0.47)) / 1006) \approx 1.08[/tex]

P-value calculation:

Since it's a right-tailed test, P-value = P(z > 1.08) = 1 - P(z ≤ 1.08).

Look up the value in the z-table to find P(z ≤ 1.08) ≈ 0.8599.

P-value ≈ 1 - 0.8599 ≈ 0.1401.

Decision:

Since the P-value (0.14) is greater than the significance level (α = 0.01), we fail to reject the null hypothesis.

There's sufficient evidence to support the claim that about 47% of the general consumer population in the United States is loyal to their chosen automobile manufacturer.

Thus, it doesn't indicate that the population proportion of consumers loyal to Chevrolet is more than 47%.

Answer:  (0.3332, 0.33341)

Step-by-step explanation:

Formula to find the confidence interval for population mean[tex](\mu)[/tex] :

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where n= Sample size

[tex]\overline{x}[/tex] = sample mean.

[tex]z^*[/tex] = Critical z-value (two-tailed)

[tex]\sigma[/tex] = population standard deviation.

As per given , we have

n= 53

[tex]\sigma=0.000507\ mm [/tex]

[tex]\overline{x}=0.3333\ mm[/tex]

The critical values for 90% confidence interval : [tex]z^*=\pm1.645[/tex]

Now , the 90 percent confidence interval for the true mean metal thickness:

[tex]0.3333\pm (1.645)\dfrac{0.000507}{\sqrt{53}}\\\\=0.3333\pm(1.645)(0.0000696)\approx0.3333\pm0.00011456\\\\=(0.3333-0.00011456,\ 0.3333+0.00011456)\\\\=(0.33318544,\ 0.33341456)\approx(0.3332,\ 0.3334)[/tex]

Hence, the 90 percent confidence interval for the true mean metal thickness. : (0.3332, 0.3334)

Answer:Test 1 < Test 4

Step-by-step explanation: Test 1 was easier so it took less time to complete the questions. As the tests continue they get harder meaning they will take more time to answer.

Answer:

the answer is a

Step-by-step explanation:

i just took the test

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

The test statistic for the main effect for factor A cannot be computed from the provided information because the mean square for error (MSE) is not given, which is required for the F statistic calculation.

The test statistic for determining whether there is a main effect for factor A is calculated as follows.

First, the mean square for factor A (MSA) is found by dividing SSA by its degrees of freedom, which for factor A is a-1. In this case, with a = 6, degrees of freedom for A is 6-1, which is 5.

Therefore, MSA = SSA / dfA = 1.05 / 5 = 0.21.

The mean square for error (MSE) is not provided, which is necessary to calculate the F statistic for factor A.

To obtain the test statistic for factor A, we need to divide MSA by MSE. Without SSE or MSE, the test statistic cannot be computed. Hence, the correct answer is c.

The test statistic cannot be computed because SSE is not given.

Answer:

(-3.0486, -0.2848)

Step-by-step explanation:

Let the number of hours per day spent using electronic media from 1999 be the first population and the number of hours per day spent using electronic media from 1999 the second population.  

We have small sample sizes [tex]n_{1} = 15[/tex] and

[tex]n_{2} = 15[/tex].

[tex]\bar{x}_{1} = 5.9333[/tex] and [tex]\bar{x}_{2} = 7.6[/tex]; [tex]s_{1} = 1.0998[/tex] and [tex]s_{2} = 1.5946[/tex].  

The pooled estimate is given by  

[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(1.0998)^{2}+(15-1)(1.5946)^{2}}{15+15-2} = 1.8762[/tex]

The 99% confidence interval for the true mean difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is given by  

[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex], i.e.,

[tex](5.9333-7.6)\pm t_{0.005}1.3697\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex]

where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (15+15-2) = 28 degrees of freedom. So

[tex]-1.6667\pm(-2.7633)(1.3697)(0.3651)[/tex], i.e.,

(-3.0486, -0.2848)

######

Using R

time1999 <- c(4, 5, 7, 7, 5, 7, 5, 6, 5, 6, 7, 8, 5, 6, 6)

time2009 <- c(5, 9, 5, 8, 7, 6, 7, 9, 7, 9, 6, 9, 10, 9, 8)

n1 <- length(time1999)

n2 <- length(time2009)

(mean(time1999)-mean(time2009))+qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)

(mean(time1999)-mean(time2009))-qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)

To find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999, we can use the formula: CI = (x1 - x2) ± Z * sqrt((s1^2/n1) + (s2^2/n2)). Using R, the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is (0.4733, 2.1933) hours.

To find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999, we can use the formula:

CI = (x1 - x2) ± Z * sqrt((s1^2/n1) + (s2^2/n2))

where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the critical value for a 99% confidence level.

Using R, we can calculate the confidence interval as follows:

The 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is (0.4733, 2.1933) hours. This means we are 99% confident that the true difference between the means falls within this interval.

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Good evening ,

Answer:

x=2

Step-by-step explanation:

Look at the photo below for the details.

:)

a.

[tex]f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}[/tex]

is a proper joint density function if, over its support, [tex]f[/tex] is non-negative and the integral of [tex]f[/tex] is 1. The first condition is easily met as long as [tex]C\ge0[/tex]. To meet the second condition, we require

[tex]\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}[/tex]

b. Find the marginal joint density of [tex]X[/tex] and [tex]Y[/tex] by integrating the joint density with respect to [tex]z[/tex]:

[tex]f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz[/tex]

[tex]\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}[/tex]

Then

[tex]\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy[/tex]

[tex]\approx\boxed{0.12886}[/tex]

c. This probability can be found by simply integrating the joint density:

[tex]\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]

[tex]\approx\boxed{0.012262}[/tex]

(a) We determined the constant C by integrating the joint density function over the entire space, finding = 1/10.

​(b) We calculated P(X≤1.375,Y≤1.5) by integrating the joint density function over the specified region, resulting in approximately 0.1286.

(c) For P(X≤1.375,Y≤1.5,Z≤1), we utilized the results from part (b) and integrated over, yielding approximately 0.1163.

The problem involves determining the constant C and calculating certain probabilities for the given joint density function f(x, y, z).

Below is the step-by-step solution:

Part (a): Finding the Constant C

First, we need to find the value of C such that the total probability is 1.

This means we need to evaluate the integral of the joint density function over the entire space.

[tex]\int \int \int C e^{-(0.5x + 0.2y + 0.1z)} \, dx \, dy \, dz = 1[/tex], where x, y, z ≥ 0

[tex]\int_{0}^{\infty} Ce^{-0.5x} \, dx = \frac{C}{0.5}[/tex]

[tex]\int_{0}^{\infty} e^{-0.5x} \, dx = \frac{1}{0.5}[/tex] from 0 to ∞ = 2C

[tex]\int_{0}^{\infty} e^{-0.2y} \, dy = \frac{1}{0.2} = 5[/tex]

[tex]\int_{0}^{\infty} e^{-0.1z} \, dz = \frac{1}{0.1} = 10[/tex]

2C * 5C * 10C = 1

100C³ = 1

C = 1 / 10

Part (b): Finding P(X ≤ 1.375 , Y ≤ 1.5)

P(X ≤ 1.375 , Y ≤ 1.5) = [tex]\int_{0}^{1.375} \int_{0}^{1.5} 0.1 e^{-0.5x} e^{-0.2y} \, dy \, dx[/tex]

[tex]\int_{0}^{1.5} e^{-0.2y} \, dy = -\frac{1}{0.2} (e^{-0.3} - 1)[/tex]

= [tex]\frac{1}{0.2} (1 - e^{-0.3}) = \frac{5}{3} (1 - e^{-0.3})[/tex]

= 5 * 0.2592

≈ 1.296

[tex]\int_{0}^{1.375} 0.1 \cdot 1.296 \cdot e^{-0.5x} \, dx = 0.1 \cdot 1.296 \cdot \left( 2 - 2 e^{-0.6875} \right)[/tex]

= [tex]0.1296 \cdot \left[ -2 e^{-0.5x} \right]_{0}^{1.375}[/tex]

= [tex]0.1296 \cdot 2 \cdot \left( 1 - e^{-0.6875} \right)[/tex]

= 0.1296 * 2 * (1 - 0.5038)

≈ 0.1296 * 2 * 0.4962

≈ 0.1286

Part (c): Finding P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1)

P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1) = [tex]\int_{0}^{1.375} \int_{0}^{1.5} \int_{0}^{1} 0.1 e^{-0.5x} e^{-0.2y} e^{-0.1z} \, dz \, dy \, dx[/tex]

[tex]\int_{0}^{1} e^{-0.1z} dz = \left[ -\frac{1}{0.1} e^{-0.1z} \right]_{0}^{1}[/tex]

= [tex](1/0.1)(1 - e^{-0.1})[/tex]

= [tex]10 (1 - e^{-0.1})[/tex]

≈ 0.905

Combining all parts:

P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1) = 0.1286 * 0.905

≈ 0.1163

The size of the sample space for grouping 10 kids into two teams is 252. The probability of Alex and Jose being on the same team is 4/9.

To calculate the sample space for randomly grouping 10 kids into two teams of five, we consider the number of ways to choose 5 kids out of 10, since choosing the first team automatically determines the second. The size of the sample space is given by the combination formula C(n, k) = n! / (k!(n - k)!), where n! represents the factorial of n. Therefore, the sample space size is C(10, 5) = 10! / (5!5!) = 252.

To find the probability that Alex and Jose end up on the same team, we need to consider two scenarios: both in team A or both in team B. Since the order they are picked doesn't matter, they are one unit, and we need to choose 3 additional members for their team from the remaining 8 kids, we use the combination formula again: C(8, 3).

There are C(8, 3) ways for them to be on the same team and this can happen in two different teams. Thus, the probability that they are on the same team is (2 * C(8, 3)) / C(10, 5).

Calculating this gives us (2 * 56) / 252, which simplifies to 112/252, and reduces to 4/9 when simplified. So, the probability is 4/9.

Answer:

0.1724 or 17.24%

Step-by-step explanation:

There are  

10+13+7 = 30 people

so there are C(30,5) combinations of 30 elements taken 5 at a time ways to select groups of 5 people.

[tex]C(30,5)=\displaystyle\binom{30}{5}=\frac{30!}{5!25!}=142,506[/tex]

There are C(10,2) combinations of 10 elements taken 2 at a time ways to select the 2 Democrats

[tex]C(10,2)=\displaystyle\binom{10}{2}=\frac{10!}{2!8!}=45[/tex]

There are C(13,2) combinations of 13 elements taken 2 at a time ways to select the 2 Republicans

[tex]C(13,2)=\displaystyle\binom{13}{2}=\frac{13!}{2!11!}=78[/tex]

There are C(7,1) combinations of 7 elements taken 1 at a time ways to select the Independent

[tex]C(7,1)=\displaystyle\binom{7}{1}=\frac{7!}{1!6!}=7[/tex]

By the Fundamental Principle of Counting, there are  

45*78*7 = 24,570

ways of making the group of 5 people with 2 Democrats, 2 Republicans and 1 independent, and the probability of forming the group is

[tex]\displaystyle\frac{24,570}{142,506}=0.1724[/tex] or 17.24%

Answer:

p value = 0.3122

Step-by-step explanation:

Given that Pedro thinks that he has a special relationship with the number 3. In particular,

Normally for a die to show 3, probability p = [tex]\frac{1}{6} =0.1667[/tex]

Or proportion p = 0.1667

Pedro claims that this probability is more than 0.1667

[tex]H_0 = P =0.1667______H_a = P>0.1667_____[/tex]

where P is the sample proportion.

b) n=30 and [tex]P = 6/30 =0.20[/tex]

Mean difference = [tex]0.2-0.1667=0.0333[/tex]

Std error for proportion = [tex]\sqrt{\frac{p(1-p)}{n} } \\=0.0681[/tex]

Test statistic Z = p difference/std error =  0.4894

p value = 0.3122

p >0.05

So not significant difference between the two proportions.

The null hypothesis is that Pedro's claim is not true, while the alternative hypothesis is that Pedro's claim is true. The P-value is approximately 0.082.

(a) The null hypothesis states that Pedro's claim is not true, so the null hypothesis is H0: p = 1/6. The alternative hypothesis states that Pedro's claim is true and he rolls a 3 more often than expected, so the alternative hypothesis is Ha: p > 1/6.

(b) To determine the P-value, we can use the binomial distribution. The probability of rolling a 3 with a fair 6-sided die is 1/6. Using the binomial distribution, we can calculate the probability of rolling a 3 six times out of 30 rolls. This gives a P-value of approximately 0.082, which is the probability of observing a result as extreme as the one obtained or more extreme, assuming the null hypothesis is true.

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Answer:

We conclude that the sonnets were written by by a certain Elizabethan poet.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8.9

Sample mean, [tex]\bar{x}[/tex] =10.2

Sample size, n = 6

Alpha, α = 0.05

Population standard deviation, σ = 2.5

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 8.88\\H_A: \mu > 8.88[/tex]

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

a) Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{10.2 - 8.9}{\frac{2.5}{\sqrt{6}} } = 1.28[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

b) We calculate the p value with the help of z-table.

P-value = 0.1003

The p-value is greater than the significance level which is 0.05

c) Since the p-value is greater than the significance level, there is not enough evidence to reject the null hypothesis and accept the null hypothesis.

Thus, we conclude that the sonnets were written by by a certain Elizabethan poet.

The z-score is 1.86 and the p-value is 0.0314. As the p-value is less than the level of significance α (0.05), we reject the null hypothesis and conclude that the new sonnets were likely written by another author.

In this statistical testing scenario for authorship of literary works, we need to find out the z-score or z test statistic and then determine the p-value to check if the new sonnets could be the works of the known Elizabethan poet or not.

For calculating the z score, you use the formula z = (x~ - μ) / (σ / √n) = (10.2 - 8.9) / (2.5/ √6) = 1.86 to two decimal places. The p-value is determined from the standard normal distribution table which for a z-score of 1.86 is 0.0314.

Given that α = 0.05, since the p-value is less than α, we reject the null hypothesis H0 (that the works were by the Elizabethan poet). Therefore, we accept the alternative hypothesis Ha (the sonnets were written by another author).

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Answer: 0.547

Step-by-step explanation:

As per given , we have

Population mean = [tex]\mu=20[/tex]

Population standard deviation= [tex]\sigma=4[/tex]

Sample size : n= 36

We assume that number of apps used per month by smartphone owners is normally distributed.

Let [tex]\overline{x}[/tex] be the sample mean.

Formula : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

The probability that the sample mean is between 19.5 and 20.5 :-

[tex]P(19.5<\overline{x}<20.5)\\\\=P(\dfrac{19.5-20}{\dfrac{4}{\sqrt{36}}}<\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{20.5-20}{\dfrac{4}{\sqrt{36}}})\\\\=P(-0.75<z<0.75)\\\\=P(z<0.75)-P(z<-0.75)\ \ [\because P(z_1<z<z_2)=P(z<z_2)-P(z<z_1)]\\\\=P(z<0.75)-(1-P(z<0.75))\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=2P(z<0.75)-1=2(0.7734)-1=0.5468\approx0.547[/tex]

[using standard normal distribution table for z]

Hence, the required probability = 0.547

To find the probability that the sample mean is between 19.5 and 20.5 for a random sample of 36 smartphone owners, calculate the z-scores for both values and find the area under the standard normal curve between those z-scores.

To find the probability that the sample mean is between 19.5 and 20.5 for a random sample of 36 smartphone owners, we need to calculate the z-scores for both values and find the area under the standard normal curve between those z-scores.

The formula for calculating the z-score is: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.

Using the given information, the z-scores for 19.5 and 20.5 are: z1 = (19.5 - 20) / (4 / √36) = -0.75 and z2 = (20.5 - 20) / (4 / √36) = 0.75.

Now, we can use a standard normal table or a calculator to find the area under the curve between -0.75 and 0.75. The probability is approximately 0.467.

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[tex]y=2x^2\implies\mathrm dy=4x\,\mathrm dx[/tex]

Then in the integral we have

[tex]\displaystyle\int_Cxy\,\mathrm dx+(x+y)\,\mathrm dy=\int_1^2(2x^3+4x(x+2x^2))\,\mathrm dx=\int_1^210x^3+4x^2\,\mathrm dx=\boxed{\frac{281}6}[/tex]

Scott needs to run 5/8 miles more.

Step-by-step explanation:

Distance covered = 11/8 miles

Total distance = 11/8 miles

Distance left to cover = Total distance - Distance covered

[tex]Distance\ left\ to\ cover= \frac{16}{8}-\frac{11}{8}\\Distance\ left\ to\ cover=\frac{16-11}{8}\\Distance\ left\ to\ cover=\frac{5}{8}[/tex]

Scott needs to run 5/8 miles more.

Keywords: distance, subtraction

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Answer: 0.36

Step-by-step explanation:

Given : Sample size of spools of wire : n= 24

The sample variance = [tex]s^2=0.13[/tex]

The the sample standard deviation should be the square root of the sample variance.

i.e. Sample standard deviation = [tex]s=\sqrt{0.13}=0.360555127546approx0.36[/tex]

Also, the sample standard deviation(s) is the best estimate of population standard deviation [tex](\sigma)[/tex].

Therefore ,  a point estimate for the population standard deviation [tex](\sigma)[/tex] of the diameter of the spools of wire. =s= 0.36

Hence, a point estimate for the population standard deviation of the diameter of the spools of wire= 0.36

There is no significant difference in mean lead concentration for the medicines.

Hypothesis means an assumption or a claim which is needed to be tested.

To test the following hypothesis:

[tex]Null\ hypothesis,H_0: \mu=17\\Alternative\ Hypothesis,\ H_1: \mu < 17[/tex]

For the given data to test the hypothesis, t statistics can be used as follows:

[tex]t=\dfrac{\bar{X}-\mu}{\sigma}[/tex]

Given, Mean, μ=17

We are required to sample mean and  σ in the excel table shown

below:

t=(17-13.05)/6.23

t=0.634

Using excel function to calculate p value of the t statistics as follows:

=T.DIST(0.634,9,1)

p=0.729

Since, p value is greater than level of significance 0.01. We fail to reject the null hypothesis and conclude that there is no significant difference in mean.

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To test the claim, calculate the mean and standard deviation of your sample data, set up your null and alternative hypotheses, calculate your test statistic t, reference a t-distribution table and compare your calculated t with the critical t-value.

To test the claim that the mean lead concentration for all such medicines is less than 17 mu​g/g, we'll use a one-sample t-test because we have one sample present and we're comparing it to a known mean. Here are the steps:

Remember, rejecting the null hypothesis supports our claim that the mean concentration is less than 17 mu​g/g. Not rejecting the null hypothesis means that our data does not support the claim that the mean concentration is less than 17 mu​g/g, but it isn't proof that the concentration exactly equals to or exceeds 17 mu​g/g.

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Answer:

Step-by-step explanation:

Hello!

To be able to resolve this kind of exercise nice and easy the first step is to determine your study variable and population parameter.

The variable is

X: daily dietary allowance for zinc for a male 50 years old or older. (mg/day)

The parameter of interest is the population mean. (μ)

You need to test if the average zinc intake is bellowed the recommended allowance, symbolically μ < 15. This will be the alternative hypothesis, it's complement will be the null hypothesis. (easy tip to detect the null hypothesis fast, it always carries the = sign)

The hypothesis is:

H₀: μ ≥ 15

H₁: μ < 15

α: 0,05

Now you have no information about the variable distribution. You need the variable to be normally distributed to study the population mean. Since the sample is large enough, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal, this way, you can use the approximate Z to make the test.

X[bar]≈N(μ;σ²/n)

Z=  X[bar]-μ ≈N(0;1)

√(σ²/n)

Z=  11.2 - 15  = -6.17

6.58/√114

The critical region of this test is one-tailed (left) the critical value is:

[tex]Z_{\alpha } = Z_{0.05} = -1.64[/tex]

If Z ≤ -1.64, you reject the null hypothesis.

If Z > -1.64, you support the null hypothesis

The calculated value is less than the critical value, so the decision is to reject the null hypothesis.

This means that the average daily zinc intake of males age 65 - 74 is below the recommended allowance.

I hope it helps!

Answer:

The cylinder is given straight away by x^2+y^2=r^2=16\implies r=4. To get the cylinder, we complete one revolution, so that 0\le\theta\le2\pi. The upper limit in z is a spherical cap determined by

x^2+y^2+z^2=144\iff z^2=144-r^2\implies z=\sqrt{144-r^2}

So the volume is given by

\displaystyle\iiint_{\mathcal D}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=0}^{z=\sqrt{144-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta

and has a value of \dfrac{128(27-16\sqrt2)\pi}3 (not that we care)

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Step-by-step explanation:

Answer:

70th percentile for the amount of time between customers entering is 30.10

Step-by-step explanation:

given data

mean =  25

to find out

What is the 70th percentile for the amount of time between customers entering Clay's store

solution

we know that here mean is

mean = [tex]\frac{1}{\lambda}[/tex]   ..................1

so here 25 =  [tex]\frac{1}{\lambda}[/tex]

and we consider time value corresponding to 70th percentile = x

so we can say

P(X < x) = 1 - [tex]e^{- \lambda *x}[/tex]      

P(X < x) = 70 %

1 - [tex]e^{\frac{-x}{25}}[/tex]  = 70 %

1 - [tex]e^{\frac{-x}{25} }[/tex]  = 0.70

[tex]e^{\frac{- x}{25} }[/tex]  = 0.30

take ln both side

[tex]\frac{-x}{25}[/tex] = ln 0.30

[tex]\frac{x}{25}[/tex]  = 1.203973

x = 30.10

70th percentile for the amount of time between customers entering is 30.10

The 70th percentile for the amount of time between customers entering Clay's store is 30

The mean of an exponential distribution is:

[tex]E(x) = \frac 1\lambda[/tex]

The mean is 25.

So, we have:

[tex]\frac 1\lambda = 25[/tex]

Solve for [tex]\lambda[/tex]

[tex]\lambda = \frac 1{25}[/tex]

[tex]\lambda = 0.04[/tex]

An exponential function is represented as:

[tex]P(x < x) = 1 - e^{-\lambda x}[/tex]

For the 70th percentile, we have:

[tex]1 - e^{-\lambda x} = 0.70[/tex]

This gives

[tex]e^{-\lambda x} = 1 - 0.70[/tex]

[tex]e^{-\lambda x} = 0.30[/tex]

Substitute [tex]\lambda = 0.04[/tex]

[tex]e^{-0.04x} = 0.30[/tex]

Take the natural logarithm of both sides

[tex]-0.04x = \ln(0.30)[/tex]

This gives

[tex]-0.04x = -1.20[/tex]

Divide both sides by -0.04

[tex]x = 30[/tex]

Hence, the 70th percentile for the amount of time between customers entering Clay's store is 30

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Answer: the fraction of the whole pie that Rebecca ate is 1/15

Step-by-step explanation:

Let x represent the size or area of the whole pie. Maria ate 1/3 of the pie. This means that the amount of pie that maria ate is 1/3 × x = x/3

Her sister,Rebecca ate 1/5 of that. This means that Rebecca ate 1/5 of the amount that Maria ate. The amount that Rebecca ate will be expressed as 1/5 × x/3 = x/15

To determine the fraction of the whole pie that Rebecca ate, we will divide the amount that Rebecca ate by the size of the whole pie. It becomes

x/15/x = x/15 × 1/x

= 1/15

Direct computation:

Parameterize the top part of the circle [tex]x^2+y^2=1[/tex] by

[tex]\vec r(t)=(x(t),y(t))=(\cos t,\sin t)[/tex]

with [tex]0\le t\le\pi[/tex], and the line segment by

[tex]\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)[/tex]

with [tex]0\le t\le1[/tex]. Then

[tex]\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)[/tex]

[tex]=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt[/tex]

[tex]=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}[/tex]

Using the fundamental theorem of calculus:

The integral can be written as

[tex]\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}[/tex]

If there happens to be a scalar function [tex]f[/tex] such that [tex]\vec F=\nabla f[/tex], then [tex]\vec F[/tex] is conservative and the integral is path-independent, so we only need to worry about the value of [tex]f[/tex] at the path's endpoints.

This requires

[tex]\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)[/tex]

[tex]\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C[/tex]

So we have

[tex]f(x,y)=-\cos x+\sin y+C[/tex]

which means [tex]\vec F[/tex] is indeed conservative. By the fundamental theorem, we have

[tex]\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}[/tex]

The line integral of (sin x dx + cos y dy) over curve C is calculated using parameterization and direct computation for the circle part and the line segment part, or alternative methods such as polar coordinates or contour integration.

The question involves evaluating the line integral of the function (sin x dx + cos y dy) along a given curve C, which in this case consists of two parts: the upper semicircle of x2 + y2 = 1 and a line segment from (−1, 0) to (2, −π). To evaluate this, we use two methods: direct computation and using Green's theorem when applicable. For the semicircle, we parameterize it by setting x = cos(θ), y = sin(θ), where θ ranges from 0 to π. The line segment can be parameterized by a straight line equation derived from the two points. The integral along the curve is computed separately for each piece, and the results are added together to find the integral over the entire curve. To evaluate the line integral in an alternate method, one might use tricks such as transforming to polar coordinates or employing complex analysis techniques such as contour integration, depending on the nature of the integral.

Answer:B

Step-by-step explanation: The answer is B. 1/6^7. Your trying to find the reciprocal of 6^-7. The way you do that is put a 1 over the 6 with a power of -7 like this (6^-7)/(1) and that's how you get your answer!

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{This equation should be somewhat}\\\huge\textbf{easy because it has like THREE steps}\\\huge\textbf{to solve it.}[/tex]

[tex]\huge\textbf{But....}[/tex]

[tex]\huge\textbf{Your equation should look like:}[/tex]

[tex]\mathsf{6^{-7}}[/tex]

[tex]\huge\textbf{Now, let's get to the steps:}[/tex]

[tex]\mathsf{6^{-7}}\\\\\mathsf{\approx \dfrac{1}{6\times6\times6\times6\times6\times6\times6}}\\\\\mathsf{\approx \dfrac{1}{6^7}}\\\\\mathsf{\approx \dfrac{1}{279,936}}[/tex]

[tex]\huge\textbf{Therefore, the closet answer to your}\\\huge\textbf{question should be:}[/tex]

[tex]\huge\boxed{\frak{Option\ B. \rm{\dfrac{1}{6^7}}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

Sample size should be atleast 55320

Step-by-step explanation:

Given that a  group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week.

population standard deviation [tex]\sigma = 6 hrs[/tex]

Std error = [tex]\frac{6}{\sqrt{n} }[/tex]

Margin of error for 95%

= [tex]1.96*\frac{6}{\sqrt{n} } <0.5\\\sqrt{n} >235.2\\n \geq 55319.04[/tex]

Since sample size is number of items it cannot be indecimal.

So we can round off to the next high integer on a safer side.

The required sample size for a 95% level of confidence, with a population standard deviation of six hours and error less than a half hour, is approximately 554 students.

The question is about finding the required sample size for a statistics project. This project involves figuring out the average amount of time students spend studying per week with a given population standard deviation, desired error, and level of confidence.

To solve this, we use the formula for sample size: n = [(Z_α/2*σ)/E]^2. Here, Z_α/2 is the Z-score corresponding to the desired level of confidence, σ is the population standard deviation, and E is the desired error.

For a 95% level of confidence, the Z-score is about 1.96 (you can find this value in Z-tables). Given σ = 6 (the population standard deviation stated in the question) and E = 0.5 (the desired error less than a half hour), we substitute these into the formula to get:

n = [(1.96*6)/0.5]^2

So, n ≈ 553.29. Because we can't have a fraction of a student, we should round this number up to the nearest whole number.

Therefore, "the required sample size is 554 students".

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Answer:

0.682 is the probability that one newborn baby will have a weight within one standard deviation of the​ mean.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  8 pounds

Standard Deviation, σ = 0.6 pounds

We are given that the distribution of weights of​ full-term newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(weight between 7.4 and 8.6 ​pounds)

[tex]P(7.4 \leq x \leq 8.6) = P(\displaystyle\frac{7.4 - 8}{0.6} \leq z \leq \displaystyle\frac{8.6-8}{0.6}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%[/tex]

[tex]P(7.4 \leq x \leq 8.6) = 6.82\%[/tex]

This could also be found with the empirical formula.

0.682 is the probability that one newborn baby will have a weight within 0.6 pounds of the mean.

Answer:

  300 g

Step-by-step explanation:

There are 1000 grams in one kilogram. Dividing 7200 grams into 24 equal parts makes each part ...

  (7200 g)/(24 persons) = 300 g/person

Each person get 300 grams.

Answer: There is sufficient evidence to support the claim that the bags are under-filled.

Step-by-step explanation:

Since we have given that

[tex]H_0:\mu=418\\\\H_a:\mu<418[/tex]

Sample mean = 411

Standard deviation = 20

n = 9

So, the test statistic value is given by

[tex]z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\z=\dfrac{411-418}{\dfrac{20}{\sqrt{9}}}\\\\\\z=\dfrac{-7}{\dfrac{20}{3}}\\\\\\z=-1.05[/tex]

At 0.025 level of significance,

critical value z = -2.306

since -2.306<-1.05

so, we will reject the null hypothesis.

Yes, there is sufficient evidence to support the claim that the bags are underfilled.