Please help me figure this one out!

Answer:

A = 36pi in^2

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

A = pi 6^2

A = 36pi in^2

Radius(r)=6 inches

Area of circle=?

Now,

Area of circle=pi r ^2

=pi (6)^2

36 pi inches^2

So the right answer is 36 pi inches^2

Hope it helps

We have been given that Mr. Collins and Ms. LaPointe are saving money to buy a new motorcycle. The total amount of money Mr. Collins will save is given by the function [tex]p(x) = 62+5x[/tex]. The total amount of money Ms. LaPointe will save is given by the function [tex]a(x) = x^2+38[/tex]. We are asked to find number of months when they have the same amount of money saved.

To solve our given problem, we will equate both equations as:

[tex]x^2+38=62+5x[/tex]

[tex]x^2-5x+38=62+5x-5x[/tex]

[tex]x^2-5x+38=62[/tex]

[tex]x^2-5x+38-62=62-62[/tex]

[tex]x^2-5x-24=0[/tex]

[tex]x^2-8x+3x-24=0[/tex]

[tex]x(x-8)+3(x-8)=0[/tex]

[tex](x-8)(x+3)=0[/tex]

[tex](x-8)=0,(x+3)=0[/tex]

[tex]x=8,x=-3[/tex]

Since time cannot be negative, therefore, after 8 months they will have the same amount of money saved.

Answer:

false

Step-by-step explanation:

Answer:

Its true

Step-by-step explanation:

I took the quiz for Intro to Careers in Finance

Answer:

True

Step-by-step explanation:

The rule is that every 2 sides' sum is greater than the other side.

So, let's test it.

9+4>11

9+11>4

11+4>9

it checks out!

Answer:

 Z = 0.8528 < 2.576

The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance

Null hypothesis is Accepted at 0.01 level of significance.

There is no significance difference between the means

Step-by-step explanation:

Given data

size of the sample 'n' = 11

mean of the sample x⁻ =20.5

Mean of the Population μ = 18.7

Standard deviation of Population σ = 7

Test statistic

                  [tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]

                  [tex]Z = \frac{20.5 -18.7}{\frac{7}{\sqrt{11} } }[/tex]

                  [tex]Z = \frac{1.8}{2.1105}[/tex]

                  Z = 0.8528

critical Value

[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} = 2.576[/tex]

The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance

Null hypothesis is Accepted at 0.01 level of significance.

There is no significance difference between the means

Answer:Answer:

B. 18.7 ± 9.7

Step-by-step explanation:

Answer:

[tex]z=\frac{256-266}{\frac{16}{\sqrt{20}}}=-2.795[/tex]    

The p value for this case can be calculated with this probability:

[tex]p_v =P(z<-2.795)=0.0026[/tex]  

Step-by-step explanation:

Information provided

[tex]\bar X=256[/tex] represent the sample mean

[tex]\sigma=16[/tex] represent the population standard deviation

[tex]n=20[/tex] sample size  

[tex]\mu_o =266[/tex] represent the value that we want to test

z would represent the statistic

[tex]p_v[/tex] represent the p value

Hypothesis to verify

We need to conduct a hypothesis in order to see if inadequate food intake should have shorter gestation times, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 266[/tex]  

Alternative hypothesis:[tex]\mu < 266[/tex]  

The statistic is given by:

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

Replacing the info given we got:

[tex]z=\frac{256-266}{\frac{16}{\sqrt{20}}}=-2.795[/tex]    

The p value for this case can be calculated with this probability:

[tex]p_v =P(z<-2.795)=0.0026[/tex]  

Answer:

0.417

Step-by-step explanation:

Total event = 9

3 cards are for win

6 cards are for not win .ie neither win

He picks two cards without replacement.

Probabilty of neither to say win

= 6/9 * 5/8

=30/72

= 0.417

Answer:

There is enough evidence to support the claim of the manufacturers that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0[/tex]

The significance level is 0.05.

The sample 1 (hybrid), of size n1=21 has a mean of 32 and a standard deviation of 6.

The sample 2 (non-hybrid), of size n2=31 has a mean of 21 and a standard deviation of 3.

The difference between sample means is Md=11.

[tex]M_d=M_1-M_2=32-21=11[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{6^2}{21}+\dfrac{3^2}{31}}\\\\\\s_{M_d}=\sqrt{1.714+0.29}=\sqrt{2.005}=1.4158[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{11-0}{1.4158}=\dfrac{11}{1.4158}=7.77[/tex]

The degrees of freedom for this test are:

[tex]df=n_1+n_2-2=21+31-2=50[/tex]

This test is a right-tailed test, with 50 degrees of freedom and t=7.77, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t>7.77)=0.0000000002[/tex]

As the P-value is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.

Answer:

.5 hour, or 30 minutes

Step-by-step explanation:

280 km / 80 kmh = 3.5 hr before limiter

280 km / (80-10) kmh = 4 hr after

so .5 hour from 3.5 to 4 hours. or 30 minutes

This is provided he matains this speed at all times.

Answer:

5   : 4

Step-by-step explanation:

4:  3.2

Multiply by 10 to get rid of the decimal

4*10 : 3.2* 10

40 : 32

Divide each side by 8 to simplify

40/8   : 32/ 8

5   : 4

Complete Question

1. Waiting times​ (in minutes) of customers at a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. Find the coefficient of variation for each of the two sets of​ data, then compare the variation.

Bank A (single lines): 6.5, 6.6, 6.7, 6.8, 7.2, 7.3, 7.4, 7.6, 7.6, 7.7

Bank B (individual lines): 4.1, 5.4, 5.8, 6.3, 6.8, 7.8, 7.8, 8.6, 9.3, 9.7

- The coefficient of variation for the waiting times at Bank A is ----- %?

- The coefficient of variation for the waiting times at the Bank B is ----- ​%?

- Is there a difference in variation between the two data​ sets?

Answer:

a

The coefficient of variation for the waiting times at Bank A is [tex]l =[/tex]6.3%

b

The coefficient of variation for the waiting times at Bank B is [tex]l_1 =[/tex]25.116%

c

The waiting time of Bank B has a considerable higher variation than that of Bank A

Step-by-step explanation:

From the question we are told that

  For Bank A  : 6.5, 6.6, 6.7, 6.8, 7.2, 7.3, 7.4, 7.6, 7.6, 7.7

 For  Bank B : 4.1, 5.4, 5.8, 6.3, 6.8, 7.8, 7.8, 8.6, 9.3, 9.7

The sample size is  n =10

The mean for Bank A is

          [tex]\mu_A = \frac{6.5+ 6.6+ 6.7+ 6.8+ 7.2+ 7.3+ 7.4+ 7.6+ 7.6+ 7.7}{10}[/tex]

          [tex]\mu_A = 7.14[/tex]

The standard deviation is mathematically represented as

      [tex]\sigma = \sqrt{\frac{\sum|x- \mu|}{n} }[/tex]

        [tex]k = \sum |x- \mu | ^2 = 6.5 -7.14|^2 + |6.6-7.14|^2+ |6.7-7.14|^2+ |6.8-7.14|^2 + |7.2-7.14|^2+ |7.3-7.14|^2, |7.4-7.14|^2+ |7.6-7.14|^2+|7.6-7.14|^2+|7.7-7.14|^2[/tex]

[tex]k = 2.42655[/tex]

    [tex]\sigma = \sqrt{\frac{2.42655}{10} }[/tex]

    [tex]\sigma = 0.493[/tex]

The coefficient of variation for the waiting times at Bank A is  mathematically represented as  

        [tex]l = \frac{\sigma}{\mu} *100[/tex]

        [tex]l = \frac{0.493}{7.14} *100[/tex]

       [tex]l =[/tex]6.3%

Considering Bank B

     The mean for Bank B is

               [tex]\mu_1 = \frac{4.1+ 5.4+ 5.8+ 6.3+6.8+ 7.8+ 7.8+ 8.6+ 9.3+ 9.7}{10}[/tex]

              [tex]\mu_1 = 7.16[/tex]

The standard deviation is mathematically represented as      

       [tex]\sigma_1 = \sqrt{\frac{\sum|x- \mu_1|}{n} }[/tex]

    [tex]\sum |x- \mu_1 | ^2 =4.1-7.16|^2 +| 5.4-7.16|^2+ |5.8-7.16|^2 + | 6.3-7.16|^2 + |6.8-7.16|^2 + | 7.8-7.16|^2 +|7.8-7.16|^2 +|8.6-7.16|^2 + |9.3-7.16|^2 +|9.7-7.16|^2[/tex]

[tex]\sum |x- \mu_1 | ^2 =32.34[/tex]

    [tex]\sigma_1 = \sqrt{\frac{32.34}{10} }[/tex]

     [tex]\sigma_1 = 1.7983[/tex]

The coefficient of variation for the waiting times at Bank B is  mathematically represented as  

        [tex]l_1 = \frac{\sigma }{\mu} *100[/tex]

        [tex]l_1 = \frac{1.7983 }{7.16} *100[/tex]

        [tex]l_1 =[/tex]25.116%

Answer:

a

Step-by-step explanation:

bc I just took the quiz☝

Answer:

  1/(t+4)²

Step-by-step explanation:

  [tex]\dfrac{t+3}{t+4}\div(t^2+7t+12)=\dfrac{t+3}{t+4}\cdot\dfrac{1}{(t+3)(t+4)}=\dfrac{(t+3)}{(t+3)(t+4)^2}\\\\=\boxed{\dfrac{1}{(t+4)^2}}[/tex]

Answer:

98.57

Step-by-step explanation:

3.5% over 100% X 102=3.43

102-3.43= 98.57

Answer:

Step-by-step explanation:

Hello!

A group of volunteers is randomly assigned to two groups:

Group 1: watch a 5-minute video explaining good strategies for completing the mazes.

X[bar]₁ = 11.98, S₁ = 8.69, n₁ = 43

Group 2: watch a 5-minute video of other people successfully completing the mazes, but with no explanation given.

X[bar]₂ = 9.15, S₂ = 7.75, n₂ = 52

After watching the videos the volunteers were given a set of pencil and paper mazes to resolve and the time, in minutes, it took them to resolve the mazes was measured.

a)

An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information. These types of studies are usually made as precursors to a more formal experimental study, to have an idea of what's to be expected from the population.

An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It is also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).

⇒ Considering these definitions, and the description of the experiment, how the volunteers were treated exactly the same except for the videos and that the assignment of the groups was random, you can classify it as an experimental study.

b.

The response variable is the one that was measured by the researchers.

X: Time it takes the volunteer to complete the paper mazes.

This variable is quantitative continuous.

The predictor variable is the variable suspected to modify the response variable:

Y: Type of video assigned to the volunteer. Categorized: "Video on good strategies to solve mazes" and "Video showing people solving mazes"

This variable is a qualitative categorical.

If you want to compare the times it takes the volunteers of both groups the best is to do so trough the population means, so the parameter of interest is: μ₁ - μ₂

The claim is that there is no difference between the times that it takes people to complete mazes after watching either video so the hypotheses are:

H₀:  μ₁ - μ₂=0

H₁:  μ₁ - μ₂≠0

c.

Assuming X[bar]₁≈N and X[bar]₂≈N (since both samples n₁ and n₂ are large enough you can approximate the distribution of the sample means using the central limit theorem)

(X[bar]₁-X[bar]₂)≈N(μ₁-μ₂;σ₁²/n₁+σ₂²/n₂)

The estimation of the variance σ₁²/n₁+σ₂²/n₂ is:

V(X)= [tex]\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} = \frac{(8.69)^2}{43} +\frac{(7.75)^2}{52}= 2.91[/tex]

Standard error= √V(X)= √2.91= 1.706

d.

(X[bar]₁-X[bar]₂)±[tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}}[/tex]

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]

(11.98-9.15)±1.96*1.706

[-0.51; 6.17]minutes

e.

The 95% confidence interval contains the zero, so using this CI and at a complementary significance level of 5%, the test is not significant, which means that there is no evidence to reject the null hypothesis.

Correct option: We do not have evidence that there is a difference in population means, because the null value is inside the 95% CI.

f.

You can make two types of errors when deciding over a hypothesis test:

Type I error: Reject the null hypothesis when the hypothesis is true.

Type II error: Fail to reject the null hypothesis when the hypothesis is false.

Since the null hypothesis wasn't rejected, there is a chance that a type II error was committed.

The correct option is:

6. It is possible that we made a Type II error because this is when you fail to reject a false null hypothesis.

I hope this helps!

Answer:

[tex]30[/tex]

Step-by-step explanation:

There are 3 outcomes out of 5 that are odd.

[tex]\frac{3}{5}[/tex]

Throw it 50 times.

[tex]50*\frac{3}{5}[/tex]

Answer:

30.

Step-by-step explanation:

The probability of throwing a 1, 3 or 5 on one throw is 3/5.

So on 50 throws the average would be  50 * 3/5 = 30 times.

Answer:

The answer is x=5.

Step-by-step explanation:

11x + 7x + 90 = 180

18x + 90 = 180

       -90    -90

= 18x = 90

 x = 5

x is 5!

Value of x is 10°.

Correct option is B.

A figure which is formed by two rays or lines that shares a common endpoint is called an angle. The word “angle” is derived from the Latin word “angulus”, which means “corner”.

Given,

a || b

cut by transversal s and t

lines b and t have angle = 90°

The first top left angle between a, s, t = 7x

The last bottom angle between s and b = 11x

a || b, so a and t will also have angle = 90°

by the figure,

180° - 11x + 90° - 7x + 90° = 180°

360° - 18x = 180°

18x = 180°

x = 10°

Hence, 10° is value of x.

Learn more about angle here:

brainly.com/question/28451077

#SPJ6

Answer:

A cup of water can hold a cup of water (or about 250 mL)

Step-by-step explanation:

anything between 1 and 4 cups should be an acceptable answer

If this helps, please consider giving me brainliest

Answer:

I'm gonna go with 250 ML

Step-by-step explanation:

One liter is a little more than 1 quart.

250 mL = 0.25 L

Answer:

P(X ≥ 125) = 0.9812

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the normal distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\lambda[/tex] is the mean in the given interval, which is the same as the variance.

Normal distribution:

When the distribution is normal, we use 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.

The Poisson can be approximated to the normal, with [tex]\mu = \lambda, \sigma = \sqrt{\lambda}[/tex]

Let μ = 2.5 every minute

This is the mean of the Poisson, so [tex]\lambda = 2.5n[/tex], in which n is the number of minutes.

P(X ≥ 125) over an hour

An hour has 60 minutes, so [tex]n = 60, \lambda = 2.5*60 = 150, \sigma = \sqrt{150} = 12.25[/tex]

Using continuity correction, this is [tex]P(X \geq 125 - 0.5) = P(X \geq 124.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 124.5. So

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

[tex]Z = \frac{124.5 - 150}{12.25}[/tex]

[tex]Z = -2.08[/tex]

[tex]Z = -2.08[/tex] has a pvalue of 0.0188

1 - 0.0188 = 0.9812

So

P(X ≥ 125) = 0.9812

The correct option is d. ∆UVW ∼ ∆UWT ∼ ∆WVT

The following information should be considered:

learn more: https://brainly.com/question/994316?referrer=searchResults

Answer:

UVW ~ ΔUWT ~ ΔWVT

Step-by-step explanation:

Answer:

f(x)=4

Step-by-step explanation:

In mathematics, exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. It can be expressed by the formula y=a(1-b)x wherein y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed.

f(x)=4(2/3)*

option D

opción D

Answer:

$11,500 was invested at 13%.

$17,500 was invested at 4%

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

[tex]E = P*I*t[/tex]

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

[tex]T = E + P[/tex]

In this question:

Loans totaling 29,000.

P was invested at 13%

29000 - P was invested at 4%.

First investment:

Principal P.

Interest 13% = 0.13.

One year, so t = 1.

So

[tex]E_{1} = P*0.13*1[/tex]

[tex]E_{1} = 0.13P[/tex]

Second investment:

Principal 29000 - P.

Interest 4% = 0.04.

One year, so t = 1.

So

[tex]E_{2} = (29000-P)*0.04[/tex]

The total interest earned for both loans was $2,195.00.

This means that [tex]E_{1} + E_{2} = 2195[/tex]

So

[tex]E_{2} = 2195 - E_{1}[/tex]

So we solve the following system:

[tex]E_{1} = 0.13P[/tex]

[tex]E_{2} = (29000-P)*0.04[/tex]

[tex]2195 - E_{1} = (29000-P)*0.04[/tex]

[tex]2195 - 0.13P = 1160 - 0.04P[/tex]

[tex]0.09P = 2195 - 1160[/tex]

[tex]P = \frac{2195 - 1160}{0.09}[/tex]

[tex]P = 11500[/tex]

$11,500 was invested at 13%.

29000 - 11500 = 17500

$17,500 was invested at 4%

Answer:

141

I got it right

Answer:

a) [tex] z = \frac{295.2-247}{62.2}=0.772[/tex]

And using the normal distribution table or excel we got:

[tex] P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006[/tex]

b) [tex] z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119[/tex]

And we can use the normal standard table or excel in order to find the probability and we got:

[tex] P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907[/tex]

Step-by-step explanation:

For this case we know that the random variable of interest is normally distributed with the following parameters:

[tex] X \sim N (\mu = 247, \sigma =62.2)[/tex]

Part a

We want to find this probability:

[tex] P(X>295.2)[/tex]

And we can use the z score formula given by:

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

Replacing we got:

[tex] z = \frac{295.2-247}{62.2}=0.772[/tex]

And using the normal distribution table or excel we got:

[tex] P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006[/tex]

Part b

We select a random sample of size n = 16 and we try to find this probability:

[tex] P(\bar X >295.2)[/tex]

And we can use the z score formula given by:

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And replacing we got:

[tex] z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119[/tex]

And we can use the normal standard table or excel in order to find the probability and we got:

[tex] P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907[/tex]

Answer:

  0

Step-by-step explanation:

Put 0 where x is and do the arithmetic.

  R(0) = 2(0) = 0

  P(R(0)) = P(0) = 3(0)

  P(R(0)) = 0

Answer:

18.84mm^2

Step-by-step explanation:

Surface area of a cylinder can be found with the following formula:

2πrh+2πr^2

r is radius

h is height

Plug our values in

2π1(2)+2π(1)^2

2π(2)+2π(1)

4π+2π

6π

6(3.14)

18.84

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

                                  P.Q. =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean nicotine content = 27.3 milligrams

            s = sample standard deviation = 2.8 milligrams

            n = sample of cigarettes = 9

            [tex]\mu[/tex] = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-3.355 < [tex]t_8[/tex] < 3.355) = 0.99  {As the critical value of t at 8 degree

                                      of freedom are -3.355 & 3.355 with P = 0.5%}  

P(-3.355 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 3.355) = 0.99

P( [tex]-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99

P( [tex]\bar X-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99

99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                                          = [ [tex]27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }[/tex] , [tex]27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }[/tex] ]

                                          = [27.3 [tex]\pm[/tex] 3.131]

                                          = [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

                                  P.Q. =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean nicotine content = 24.9 milligrams

            s = sample standard deviation = 2.6 milligrams

            n = sample of cigarettes = 9

            [tex]\mu[/tex] = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-2.896 < [tex]t_8[/tex] < 2.896) = 0.98  {As the critical value of t at 8 degree

                                       of freedom are -2.896 & 2.896 with P = 1%}  

P(-2.896 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.896) = 0.98

P( [tex]-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98

P( [tex]\bar X-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98

98% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                                          = [ [tex]24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }[/tex] , [tex]24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }[/tex] ]

                                          = [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

Answer:

Step-by-step explanation:

Let x represent the random variable representing the scores in the exam. Given that the scores are normally distributed with a mean of 40 and a standard deviation of 5, the diagram representing the curve and​ the position of the​ mean, the mean plus or minus one standard​ deviation, the mean plus or minus two standard​ deviations, and the mean plus or minus three standard deviations is shown in the attached photo

1 standard deviation = 5

2 standard deviations = 2 × 5 = 10

3 standard deviations = 3 × 5 = 15

1 standard deviation from the mean lies between (40 - 5) and (40 + 5)

2 standard deviations from the mean lies between (40 - 10) and (40 + 10)

3 standard deviations from the mean lies between (40 - 15) and (40 + 15)

b) We would apply the probability for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 40

σ = 5

the probability that a randomly selected score will be greater than 50 is expressed as

P(x > 50) = 1 - P( ≤ x 50)

For x = 50,

z = (50 - 40)/5 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.98

P(x > 50) = 1 - 0.98 = 0.02

Answer:

A) 2

Step-by-step explanation:

Start off with the given information. The question states that the x-int. is 4, so you should recognize that there is a point at (4,0). Plug the point into the equation.

k(4) + 2(0) + 8 = 0

Now simplify the equation.

4k + 0 + 8 = 0

Isolate the variable, make sure it's on its own side.

4k = 8

Now get the k by itself to solve the equation. Divide both sides by 4.

k = 2