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3 years ago

What is the product of(-1/4)x(-3/7)?write your answer in the simplest fraction form.

Mathematics

1 answer:

Murrr4er [49]3 years ago

8 0

Is the 'x' a variable or multiplication ???

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aniked [119]

Answer:

39.3%

Step-by-step explanation:

5 0

3 years ago

Solve 1 And 2 Step Equations.

masha68 [24]

Answer:

ksbdksvudkdbckx skxid SORRRYYYYY BUT I NEED POINTSSSSS :(

7 0

3 years ago

A hypothetical square grows so that the length of its diagonals are increasing at a rate of 8m/min. How fast is the area of the

ad-work [718]

Answer: The area of the square is increasing at a rate of 90.4 m2/min (square meters/minute)

Step-by-step explanation: Please see the attachments below

7 0

3 years ago

Read 2 more answers

Use the ladder method to find the factored version of this expression. (3x-18)

Vera_Pavlovna [14]

Answer:

-51

Step-by-step explanation:

cheggs acc

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