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3 years ago

Question 12 (5 points)

Mathematics

1 answer:

NemiM [27]3 years ago

6 0

Answer:

D) If WX = XY and XY = YZ, then WX = YZ.

Step-by-step explanation:

The transitive property states that if a=b and b=c, then a=c. So, if two different values equal to the same, separate value, then they are also congruent. The answer choice, d, represents this idea because WX and YZ both equal XY and are shown to be equal.

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With a height of 68 in, Nelson was the shortest president of a particular club in the past century. The club presidents of the

Ivahew [28]

Answer:

a. The positive difference between Nelson's height and the population mean is: $\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: $\\ z = -1.1739 \approx -1.174$ (Nelson's height is below the population's mean 1.174 standard deviations units).

d. Nelson's height is usual since $\\ -2 < -1.174 < 2$ .

Step-by-step explanation:

The key concept to answer this question is the z-score. A z-score "tells us" the distance from the population's mean of a raw score in standard deviation units. A positive value for a z-score indicates that the raw score is above the population mean, whereas a negative value tells us that the raw score is below the population mean. The formula to obtain this z-score is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

$\\ z$ is the z-score.

$\\ \mu$ is the population mean.

$\\ \sigma$ is the population standard deviation.

From the question, we have that:

Nelson's height is 68 in. In this case, the raw score is 68 in $\\ x = 68$ in.
$\\ \mu = 70.7$ in.
$\\ \sigma = 2.3$ in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson's height and the mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

$\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

That is, the positive difference is 2.7 in.

b. How many standard deviations is that [the difference found in part (a)]?

To find how many standard deviations is that, we need to divide that difference by the population standard deviation. That is:

$\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174$

In words, the difference found in part (a) is 1.174 standard deviations from the mean. Notice that we are not taking into account here if the raw score, x, is below or above the mean.

c. Convert Nelson's height to a z score.

Using formula [1], we have

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}$

$\\ z = \frac{-2.7\;in}{2.3\;in}$

$\\ z = -1.1739 \approx -1.174$

This z-score "tells us" that Nelson's height is 1.174 standard deviations below the population mean (notice the negative symbol in the above result), i.e., Nelson's height is below the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider "usual" heights to be those that convert to z scores between minus2 and 2, is Nelson's height usual or unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

$\\ -2 < z_{Nelson} < 2$

$\\ -2 < -1.174 < 2$

Then, Nelson's height is usual according to that statement.

7 0

3 years ago

Plz help me quick!!!!!!!!!!!!!!!

Sedbober [7]

Answer:

The answer is 1 over 5

Step-by-step explanation:

10 divided by 5 equals 2. That is your width. The same rule applies to your height. So you would get 10 over 50 which simplif equals 1 over 5.

6 0

4 years ago

Lauren spent 24 hours on the computer last week. She spent 2/3 of the time doing homework. How much time did Lauren spend doing

-BARSIC- [3]

Answer:

If Lauren spend 2/3 of her 24-hour computer time doing homework, this means she spent 16 hours doing homework. Each third of the 24 hours is 8 hours, so two thirds would be 16 hours.

6 0

3 years ago

Read 2 more answers

PLEASE SEE PICTURE

klemol [59]

I think is the correct answer

3 0

3 years ago

Gina has a meeting in 45 minutes. She is driving at a rate of 30 mi/hr. If she is 24 mi away , will she arrive on time? If not,

garri49 [273]

If she is driving at 30 miles a hour it takes two minutes to cover a mile 60 divided by 30

if she is driving at 2 minutes a mile and is going 24 miles it will take 48 minutes

2*24 =48

solution she is a slow driver she needs to pick up the pace or she will be 3 minutes late to work

45-48=-3

8 0

4 years ago