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BigorU [14]
2 years ago
8

Find the area of the shaded region

Mathematics
1 answer:
xxTIMURxx [149]2 years ago
3 0

Answer:

25.12

Step-by-step explanation:

8π = P

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3 years ago
NEED HELP ASAP!!!!!!!!!!!!!!!!
FromTheMoon [43]

Answer:

6:8 and 9:12

Step-by-step explanation:

As long as you do the same thing to both parts of the ratio, it's the same ratio

5 0
3 years ago
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A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Const
Ludmilka [50]

Answer:

0.609 \leq \sigma \leq 0.772  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation  

s=0.68 represent the sample standard deviation  

\bar x =98.90 represent the sample mean  

n=98 the sample size  

Confidence=90% or 0.90  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .  

2) Calculating the confidence interval  

The confidence interval for the population variance is given by the following formula:  

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}  

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:  

df=n-1=98-1=97  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:  

\chi^2_{\alpha/2}=120.990  

\chi^2_{1- \alpha/2}=75.282  

And replacing into the formula for the interval we got:  

\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}  

0.371 \leq \sigma^2 \leq 0.596  

Now we just take square root on both sides of the interval and we got:  

0.609 \leq \sigma \leq 0.772  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

3 0
3 years ago
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