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3 years ago

The question is in the picture.

Mathematics

1 answer:

Bess [88]3 years ago

3 0

Answer:

Henry

Step-by-step explanation:

Henry is wrong before in the second step he didn't distribute properly. He multiply 5 · x but didn't multiply 5 · -2

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Which situation is best modeled by the inequality a ≥ 16?You must be at least 16 years old to get a driver’s license.

HACTEHA [7]

C, you must be older than 16 to get a driver's license.

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4 years ago

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What is the median !!! help!! alegbra 2. marking brainliest

dedylja [7]

Answer:

The answer is 8.5

Step-by-step explanation:

Just list all the numbers and cross them out one by one.

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3 years ago

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A) 2 x -7 = b) -3 x 4 = c) -5 % -6 =

erastovalidia [21]

Answer:

a)2×-7= -14

b)-3×4= -12

c)-5%+6= -6.05

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3 years ago

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A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.

Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean $(\mu)$ = 500 and a standard deviation $(\sigma)$ = 100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = $\dfrac{Y-\mu}{\sigma}$

Z-score = $\dfrac{625-500}{100}$

⇒ Z-score = $\dfrac{125}{100}$

⇒Z-score =1.25

Therefore , the required z-score = 1.25

6 0

3 years ago

In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The confidence interval of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the level of significance $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the z-score corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative standard normal table, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a normal distribution. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

7 0

4 years ago