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3 years ago

What is the Best approximation for the perimeter of a Semi circle with a diameter of 64 m

Mathematics

1 answer:

VLD [36.1K]3 years ago

3 0

Answer:

164.848 m

Step-by-step explanation:

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Factor <br> 5(x^2n-1)·(2x^3n 1)^2

nikdorinn [45]

I think I may be wrong check

5(x^2n-1)×(2x^3n-1) ^2

=20n3 x8 -20n2 x5 +20n x3 +20n x3 +5n x2-5

3 0

3 years ago

3-4y=5x-6. The answer is (-5x^4-9)/4. Please explain how to get this.

xenn [34]

Answer:

The direction of mathematics teacher is study about the mathematics teacher is study about it is by step instructions

5 0

2 years ago

What is 183 for 4 as a unite rate

olganol [36]

Answer:

45.75

Step-by-step explanation:

183/4=45.75.

3 0

3 years ago

Read 2 more answers

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

Which equation can be used to find the unknown length, a, in this triangle?

Solnce55 [7]

Using the Pythagorean Theorem a^2 + b^2 = C^2

a and b are the sides and C is the Hypotenuse.

The picture you have sides of a and 15 and the hypotenuse is 17

The formula becomes a^2 + 15^2 = 17^2

7 0

4 years ago

Read 2 more answers