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3 years ago

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Mathematics

2 answers:

icang [17]3 years ago

8 0

The measure of angle MKJ is equal to the sum of angles MKL and JKL by the angle addition postulate. that means
2x+10 + 3x-5=80. solve for x...

5x+5=80
5x=75
x=15

now plug 15 in n for x in both angles. you'll see that the measure of both angles is 40. since segment KL cuts the big angle exactly in half, it is the angle bisector. your checked answer is the right one.

Rama09 [41]3 years ago

6 0

We know that angle MKJ is comprised of angle MKL and angle LKJ. That means if we add MKL and LKJ, we should get 80 degrees, which is the measure of angle MKJ.

$MKL + LKJ= MKJ \implies \\ 2x+10+3x-5=80 \implies \\ 5x+5=80 \implies\\ 5x=75 \implies \\ x=15$

So, we know that our x is 15. That is not enough to tell whether KL is an angle bisector, because we have to evaluate both MKL and LKJ with x=15, so:

$MKL=2(15)+10 = 40\\ LKJ=3*15-5=40$

So we see that these two angles are actually bisectors, and the third question best describes this phenomenon.

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Read 2 more answers

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

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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.

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3 years ago