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photoshop1234 [79]
3 years ago
6

What is the number if: 50% of 25% of the number is 96 Please answer fast! :)

Mathematics
2 answers:
Over [174]3 years ago
5 0

Answer:

768

Step-by-step explanation:

First multiply by two

Then multiply by four

andrey2020 [161]3 years ago
4 0

Answer: 768

Step-by-step explanation:

Because it’s 50% of 25% of 96, you may know that 100 divided by 25=4 So it’s kinda the same concept with this, you first can do 96*4=384, then do 384*2=768. You would multiply by 2 because 100/2=50, and what not. I know this is the same thing as the other person but whatevvvv

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Step-by-step explanation:

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An insurance company states that it settles 85% of all life insurance claims within 30 days. A consumer group asks the state ins
BlackZzzverrR [31]

Using the z-distribution, it is found that since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence that the proportion is below 85%, that is:

H_0: p \geq 0.85

At the alternative hypothesis, it is tested if there is enough evidence that the proportion is below 85%, that is:

H_0: p < 0.85

<h3>What is the test statistic?</h3>

The test statistic is given by:

z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}

In which:

  • \overline{p} is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

For this problem, the parameters are:

p = 0.85, n = 250, \overline{p} = \frac{203}{250} = 0.812

Hence the test statistic is:

z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}

z = \frac{0.812 - 0.85}{\sqrt{\frac{0.85(0.15)}{250}}}

z = -1.68

<h3>What is the p-value and the conclusion</h3>

Using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, and z = -1.68, the p-value is of 0.0465.

Since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

More can be learned about the z-distribution at brainly.com/question/16313918

#SPJ1

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