Pls help me with this

Which of the following fractions is the largest? A. 10/13 B. 11/14 C. 14/15 D. 17/18

larisa [96]

The one with the highest numerator and denominator. 17/18

7 0

3 years ago

Read 2 more answers

Construct the confidence interval for the population mean. c=. 95 x=6. 4 o=. 3 and n=42.

Damm [24]

The confidence interval is (63.909, 64.0907).

The confidence interval of a population mean is found using the formula:

CI = x ± z(σ/√n), where CI is the confidence interval, x is the mean, z is the z-score corresponding to the confidence level, σ is the standard deviation, and n is the population size.

In the question, we are given the confidence level, c = 95%.

Z-score corresponding to this, z = 1.96.

The mean of the population, x = 6.4.

The standard deviation of the population, σ = 0.3.

The population size, n = 42.

Thus, using the formula, the confidence interval is:

CI = 6.4 ± 1.96(0.3/√42),

or, CI = 6.4 ± 0.09073037.

Thus, the confidence interval is (64 - 0.09073037, 64 + 0.09073037), or, (63.909, 64.0907).

Learn more about the confidence interval at

brainly.com/question/17097944

#SPJ4

3 0

2 years ago

Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records

madreJ [45]

Answer:

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

Step-by-step explanation:

Previous concepts

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".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

3 0

3 years ago

I need help please help me

Leviafan [203]

Is that the Algebra work/practice book? If so, go into algebra nation and they have videos where they go step by step through the questions. Hope this helps?

5 0

3 years ago

Write an explicit formula for an, the nth term of the sequence 4, -24, 144, ....

nadezda [96]

Answer:

Tn = 4(-6)^n-1

Step-by-step explanation:

Write an explicit formula for an, the nth term of the sequence 4, -24, 144, ....

The sequence is a geometric sequence

Tn = ar^n-1

a is the first term

a = 4

r = -24/4 =144/-24

r = -6

Substitute

Tn = 4(-6)^n-1

3 0

2 years ago