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PSYCHO15rus [73]
3 years ago
10

Please help mesimplify

a1" title=" \sqrt{32 - \sqrt{50} } " alt=" \sqrt{32 - \sqrt{50} } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
5 0

Answer:

\sqrt{32-5\sqrt{2} }

Step-by-step explanation: I hope it helps :)

\sqrt{32-\sqrt{50} } \\\\simplify \: \sqrt{50} \:to \:5\sqrt{2}  \\\\= \sqrt{32-5\sqrt{2} } \\\\Decimal Form: 4.992888

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9+3.5g=11-0.5g solve for g
Tanzania [10]

Answer:

g = 1/2

Step-by-step explanation:

Step 1: Add 0.5g to both sides

9 + 4g = 11

Step 2: Subtract 9 from both sides

4g = 2

Step 3: Divide by 4 on both sides

g = 2/4

Step 4: Simplify

g = 1/2

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AnnyKZ [126]

Answer:

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Step-by-step explanation:

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While rolling, a wheel with a diameter of 19/2 ft makes 133 revolutions. Find the distance that is traveled by the wheel. Use 22
mojhsa [17]
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3 years ago
The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.
valina [46]

Answer:

A sample of 179 is needed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.85}{2} = 0.075

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.075 = 0.925, so Z = 1.44.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

A previous study found that for an average family the variance is 1.69 gallon?

This means that \sigma = \sqrt{1.69} = 1.3

If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?

A sample of n is needed, and n is found for M = 0.14. So

M = z\frac{\sigma}{\sqrt{n}}

0.14 = 1.44\frac{1.3}{\sqrt{n}}

0.14\sqrt{n} = 1.44*1.3

\sqrt{n} = \frac{1.44*1.3}{0.14}

(\sqrt{n})^2 = (\frac{1.44*1.3}{0.14})^2

n = 178.8

Rounding up

A sample of 179 is needed.

7 0
3 years ago
Frank has a student loan of $55,780.
LenaWriter [7]

Answer: $6197.77 would be the right answer

hope it helps :]

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