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3 years ago

How do you solve this.

Mathematics

1 answer:

malfutka [58]3 years ago

3 0

Does it give you a formula If so what is it so I can help you

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A homeowner measured her rectangular backyard as having a length of 40 yards and a width of 25 yards. Her measurements were accu

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Step-by-step explanation:

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3 years ago

What is the missing constant term in the perfect square that starts x^2+10x?

max2010maxim [7]

<span>x^2+10x + 25 = (x + 5)^2

missing = 25

hope it helps</span>

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3 years ago

Read 2 more answers

Aubrey and Mariah are making friendship bracelets. Aubrey has three times more beads than Mariah. The equation below show this r

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So if the main question is 9q x 3 it would equal 27q

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3 years ago

Which digit is in the ten thousandths place in 4,6037?

Iteru [2.4K]

Answer:

Step-by-step explanation:

4- Ones

6- tenths

0-hundredths

3-thousandths

7-ten-thousandths

Hope this helps! :)

6 0

3 years ago

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:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago