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djyliett [7]
3 years ago
12

Explain why Josue is correct?

Mathematics
1 answer:
alexgriva [62]3 years ago
4 0
Because she did the math :)
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I don’t understand this I need help please
scoray [572]
The answer is B 2+(-35)
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HELP ASAP EXTRA POINTS <br> -42-6n=-30 solve for n.
xz_007 [3.2K]

the answer to your question is n = -2

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83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi
arlik [135]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

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

2.127 \leq \sigma \leq 2.757

Step-by-step explanation:

Information given

\bar X=32.1 represent the sample mean

\mu population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

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 degrees of freedom given by:

df=n-1=8-1=7

The confidence level 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(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

\chi^2_{\alpha/2}=104.139

\chi^2_{1- \alpha/2}=62.132

The confidence interval is given by:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

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

2.127 \leq \sigma \leq 2.757

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3 years ago
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Sav [38]

Answer:

The answer is A.16

Step-by-step explanation:

2.5/0.4= 6.25

100/6.25= 16

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Which algebraic expression is a polynomial?
Veronika [31]

Answer:

the second one

Step-by-step explanation:

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3 years ago
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