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attashe74 [19]
3 years ago
12

Solve a=2pir^2 2pirh for pi

Mathematics
1 answer:
MrRa [10]3 years ago
7 0
A = 2\pi r^{2} 2\pirh   |Divide by 4hr^{3}

\pi ^{2} = \frac{a}{4h r^{3} }          |Square root on both sides

\pi = \sqrt{ \frac{a}{4h r^{3} } }

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Describe the effect on the area of a circle when the radius is doubled
galina1969 [7]

Answer:

The original area would be multiplied by the scale factor of 2² to get the new area.

Step-by-step explanation:

If you double the radius lets so lets do an example

(a=πr²) so

if x is the radius the area would be x²π      

double the radius so 2x then the area would be

(2x)²π so 4x²π

4x²π/x²π= 4 so its 4 times greater or

D the last answer or The original area would be multiplied by the scale factor of 2² to get the new area.

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What is the value of x in the equation 8x+2=4x
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Answer:

-0.5 is the answer

 

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Anybody know the answer?
mixas84 [53]
In a circle, the measure of an inscribed angle is half the measure of the intercepted arc ⇒ x = 40/2 = 20°
8 0
3 years ago
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

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You will make 11588 mor dollars for the hourly job
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