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alexandr1967 [171]
3 years ago
15

What is the value of the expression?

Mathematics
1 answer:
Anni [7]3 years ago
5 0
The answer is d) that is the answer.

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Trials in an experiment with a polygraph include results that include cases of wrong results and cases of correct results. Use a
hram777 [196]

Answer and Step-by-step explanation:

This is a complete question

Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80​% of the time. Identify the null​hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

The computation is shown below:

The null and alternative hypothesis is

H_0 : p = 0.80

Ha : p < 0.80

\hat p = \frac{x}{ n} \\\\= \frac{74}{97}

= 0.7629

Now Test statistic = z

= \hat p - P0 / [\sqrtP0 \times (1 - P0 ) / n]

= 0.7629 - 0.80 / [\sqrt(0.80 \times 0.20) / 97]

= -0.91

Now

P-value = 0.1804

\alpha = 0.01

P-value > \alpha

So, it is Fail to reject the null hypothesis.

There is ample evidence to demonstrate that less than 80 percent of the time reports that these polygraph findings are accurate.

5 0
3 years ago
Hello can you help me the question is:write two equivalent fractions?
MakcuM [25]

Answer:

Step-by-step explanation:

1)8/12=12/18

2)12/20=18/30

3)4/16=6/24

6 0
3 years ago
I need help with this
Alex73 [517]

Answer: the third one

Step-by-step explanation:

3 0
3 years ago
HELP IM ON THE LAST ONE ILL GIVE BRAINLIEST
xxMikexx [17]

Answer:

this is hard what grade is this for

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Use linear approximation to approximate √25.3 as follows.
Sophie [7]

The idea is to use the tangent line to f(x)=\sqrt x at x=25 in order to approximate f(25.3)=\sqrt{25.3}.

We have

f(x)=\sqrt x\implies f(25)=\sqrt{25}=5

f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}

so the linear approximation to f(x) is

L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92

Hence m=\frac1{10} and b=\frac92.

Then

f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}

4 0
3 years ago
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