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3 years ago

Which statement about histograms is true?

Mathematics

2 answers:

sattari [20]3 years ago

8 0

Answer:

Histograms can be used to exhibit the shape of distributions.

Step-by-step explanation:

kondor19780726 [428]3 years ago

3 0

Answer: Histograms can be used to exhibit the shape of distributions.

Step-by-step explanation:

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15 points and crown <br> solve by factoring (x+6)(2x-5)=0<br> I am confused

Reika [66]

Remember, in order for this equation to be true either one of the polynomials has to equal 0
So:

-> x+6 = 0

x=-6

-> 2x-5 = 0

2x= 5

x= 5/2

So the answer is x=-6 and x= 5/2.

Hope this helps!

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3 years ago

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son4ous [18]

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3 years ago

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marshall27 [118]

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3 years ago

Question 7 Unsaved A couch cost $825. If the state has a 10% sale's tax. What was the price of the couch before taxes? Question

Lerok [7]

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This is what I did :
(1- .10) (825)
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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

andrew-mc [135]

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

7 0

3 years ago