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3 years ago

5

"a survey of 700 non fatal accidents showed that 183 involved faulty equipment. find the point estimate for p, the population pr

oportion of accidents that involved faulty equipment. round to 3 decimal places."

1 answer:

Mrac [35]3 years ago

6 0

Confused as mess. Need help

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"take a number x, double it and then take that result away from 10"<br><br>Write as an expression

Yuki888 [10]

Word Problem:

Take a number, <em>x</em>, double it and then take that result away from 10.

Expression:

$10-(2x)$

8 0

3 years ago

Look at the number line below.

Paladinen [302]

? -2 -1 0 1

-1-2=-3

The answer is -3

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1 year ago

Which of the following expressions has a coefficient of 6 and a constant of 9?

Tcecarenko [31]

It would be C because a coefficient is always with a variable and 6 is with a variable which makes it a coefficient

Hope this helps

Have a great day/night

6 0

3 years ago

Read 2 more answers

If x = 8 units and h = 5 units, then what is the area of the triangle shown above?<br> A.

lora16 [44]

Answer:

1/2(bh)

A=20

Step-by-step explanation:

6 0

3 years ago

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a

Vikki [24]

Answer:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

Step-by-step explanation:

Information given:

$n=131$ represent the sample size

$\hat p=0.81$ represent the estimated proportion

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.326$

And replacing into the confidence interval formula we got:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

6 0

3 years ago