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2 years ago

13

Which one should I choose.

1 answer:

Mice21 [21]2 years ago

7 0

Answer: 9

54 ÷ 2

27

3x = 27

x = 9

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Points X and are endpoints of a diameter of OW. Point Z is Points X and are endpoints of (Tother point on the circle. Find the p

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<u>Corrected Question</u>

Points X and Y are endpoints of a diameter of Circle W. Point Z is another point on the circle. Find the probability that $\angle$ XZY is a right angle.

Answer:

Probability=1

Step-by-step explanation:

<u>Theorem</u>

If an Inscribed angle intercepts a semicircle, the angle is a right angle.

Given that X and Y are endpoints of a diameter of Circle W and point Z is on the circle's circumference.

I have prepared a diagram which is attached.

Then, $\angle$ XZY is an angle which intercepts a semicircle.

By the theorem above, $\angle$ XZY is a right angle.

Therefore, the probability that $\angle$ XZY is a right angle =1

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3 years ago

Stress at work: In a poll conducted by the General Social Survey, 81% of respondents said that their jobs were sometimes or alwa

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Answer:

We are given that 81% of respondents said that their jobs were sometimes or always stressful.

So, p = 0.81

(a) Approximate the probability that 150 or fewer workers find their jobs stressful.

x = 150

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{150}{175} =\widehat{p}$

$0.8571=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8571-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=1.5882$

Refer the z table

So, P(z<150)=0.9429

So, the probability that 150 or fewer workers find their jobs stressful is 0.9429

b) Approximate the probability that more than 140 workers find their jobs stressful.

x = 140

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{140}{175} =\widehat{p}$

$0.8=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=-0.3372$

P(z<140) = 0.3707

P(z>140)=1-P(z<140)=1-0.3707=0.6293

So, the probability that more than 140 workers find their jobs stressful is 0.6293

c) Approximate the probability that the number of workers who find their jobs stressful is between 146 and 150 inclusive.

x = 146

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{146}{175} =\widehat{p}$

$0.8342=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8342-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=0.816$

From part a) P(z<150)=0.9429

So,P(z<150)-P(z<146)=0.9429 - 0.8342=0.1087

Hence the probability that the number of workers who find their jobs stressful is between 146 and 150 inclusive is 0.1087

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3 years ago