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3 years ago

The formula to calculate the area of a circle from its circumference is: A =c2/4π

Mathematics

1 answer:

densk [106]3 years ago

4 0

Answer:

(if its a true or false question) True

Step-by-step explanation:

We know-

Area of a circle =

A=(π)r2

Circumference of a circle =

C=2(π)r

Where pi is a constant and r is the radius of the circle.

Using these two formulas we can express A in terms of C as follows:

c2=[2(π)r]2

⇒ c2=4[(π)2]r2 ⇒ c2=4(π)[(π)r2]

As (π)r2=A ⇒c2=4(π)A

Therefore:

A=c2/4(π)

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P{lease help me solve this. Will give brainliest. I cant figure out what to do here

bulgar [2K]

Answer:

Step-by-step explanation:

x=81

63+36+x=180

99+x=180

99+81=180

y and z both equal 81

13+y+z=180

180-13=167

167/2 is 83.5

13+83.5+83.5

13+167

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4 years ago

And Percent na

Alenkasestr [34]

Here's link to the answer:

tinyurl.com/wpazsebu

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3 years ago

3/5 of the pets in the animal shelter are dogs. There are 15 dogs at the shelter. How many animals are not dogs? HINT: Draw a ba

iren2701 [21]

3/5 of fifteen is 9 because that's the same as saying (3/5)*15 = (45/5) = 9. Since 3/5 animals are dogs, 2/5 aren't, and 15 - 9 leaves 6 animals that aren't dogs.

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3 years ago

Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from p

V125BC [204]

Answer:

best estimate = Xd[bar]=4.80

margin of error = 8.66

The 99% confidence interval is -3.86 to 13.46

test statistic = -1.86

p-value = 0.0526

Decision: Reject the null hypothesis.

At the 5% significance level, you can conclude that the population mean of the difference between treatment 1 and treatment 2 is less than zero.

Step-by-step explanation:

Hello!

a) 99% CI

Using d=X₁-X₂ to determine the study variable Xd: the difference between treatment 1 and treatment 2.

Assuming that this variable has an approximately normal distribution: Xd≈N(μd;σ²d)

To calculate the sample mean and standard deviation you have to calculate the difference between the values of both treatments first.

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5

22-18= 4 ; 27-29= -2 ; 32-25= 7; 26-20= 6 ; 29-20= 9

n= 5

Xd[bar]= ∑X/n= 24/5= 4.80

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/4*[186-(24²)/5]= 17.7

Sd= 4.21

The parameter of interes is the population mean od the difference, μd

The best estimate for this parameter is the sample mean, Xd[bar]=4.80

Using the t-distribution, the formula for the Confidence Interval is

Xd[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$

Where the margin of error is:

$t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$ = $t_{4;0.995}$ * $\frac{Sd}{\sqrt{n} }$ = 4.604* $\frac{4.21}{\sqrt{5} }$ = 8.66

99% CI [-3.86; 13.46]

b) 5% Hypothesis test

The variable of interest is defined d=X₁-X₂; Xd: the difference between treatment 1 and treatment 2. Xd≈N(μd;σ²d)

The statistic hypotheses are:

H₀: μd = 0

H₁: μd < 0

α: 0.05

The statistic to use for this test is:

$t_{H_0}= \frac{X_d[bar]-Mu_d}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

As before you have to calculate the difference between the observation for each case and then the sample mean and standard deviation:

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5 ; Case 6 ; Case 7 ; Case 8

18-18= 0; 12-19= -7; 11-25= -14; 21-21= 0; 15-19= -4; 11-14=-3; 14-15= -1; 22-20= 2

n= 8

Xd[bar]= ∑X/n= -27/8= -3.38

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[275-(-27²)/8]= 26.27

Sd= 5.13

$t_{H_0}= \frac{-3.38-0}{\frac{5.13}{\sqrt{8} } }= -1.86$

This test is one-tailed to the left, which means that you will reject the null hypothesis to small values of t, the p-value of the test has the same direction as the rejection region, this means that it is one-tailed to the left and you can calculate it as:

P(≤-1.86)= 0.0526

The decision rule using the p-value is:

If p-value > α, do not reject the null hypothesis.

If p-value ≤ α, reject the null hypothesis.

The p-value is greater than the significance level so the decision is to reject the null hypothesis.

I hope it helps!