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Juliette [100K]
3 years ago
12

Abby took an 8-question multiple-choice quiz. Suppose

Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0
It is 0.0623 hope it helps
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Answer ASAP<br><br><br><br> Find the area of each sector. Round your answer to the nearest tenth
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3.14*18/4= 14.13
14.1 is your answer
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3 years ago
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4x+2y=-24<br> -4x+y = 12
allsm [11]

Answer:

Step-by-step explanation:

                                4x+2y = -24    -------------(i)

                              <u>  -4x+y =  12    -</u>------------(ii)

add (i) &(ii)                      3y = -12

                                         y = -12/3

                                         y = -4

Put y = (-4) in (i)

4x + 2*(-4) = -24

4x - 8 = -24

4x = -24 + 8

4x = -16

x = -16/4

x = -4

6 0
3 years ago
A company uses 4 pounds of resource 1 to make each unit of X1 and 3 pounds of resource 1 to make each unit of X2. There are only
liraira [26]

Answer:

b)

Step-by-step explanation:

4X1 + 3X2 less than or equal to 150

5 0
3 years ago
Write the number 0.5 in integers form a/b
Alja [10]
The answer is 1/2 in fraction form because 0.5 is half of 1
5 0
3 years ago
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Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a
Vikki [24]

Answer:

0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730

0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890

And the confidence interval would be:

0.730 \leq \p \leq 0.890

Step-by-step explanation:

Information given:

n=131 represent the sample size

\hat p=0.81 represent the estimated proportion

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.326

And replacing into the confidence interval formula we got:

0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730

0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890

And the confidence interval would be:

0.730 \leq \p \leq 0.890

6 0
3 years ago
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