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Vikki [24]
3 years ago
10

Evaluate (solve) the power: 2^9

Mathematics
2 answers:
Gnesinka [82]3 years ago
4 0

Answer:

512

Step-by-step explanation:

{2}^{9}  \\ 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\ 4 \times 4 \times 4 \times 4 \times 2 \\ 16 \times 16 \times 2 \\ 256 \times 2 \\ 512

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em>

<em>Can</em><em> </em><em>I</em><em> </em><em>have</em><em> </em><em>the</em><em> </em><em>brainliest</em><em> </em><em>please</em><em>?</em>

german3 years ago
3 0

Answer:

The corrext answer is, 512.

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Answer:\frac{1}{4}

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Given

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=\frac{^6C_1}{^9C_1}\times \frac{^3C_1}{^8C_1}

=\frac{6}{9}\times \frac{3}{8}

=\frac{1}{4}

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Three more than the product of two and a number x
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2x+3

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4 years ago
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The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T
Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

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{1-0.95}{2} = 0.025

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

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

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. In this problem, \sigma = 0.25

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

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

0.1 = 1.96*\frac{0.25}{\sqrt{n}}

0.1\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.1}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}

n = 24.01

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

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

0.06 = 1.96*\frac{0.25}{\sqrt{n}}

0.06\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.06}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}

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Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

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

0.05 = 1.96*\frac{0.25}{\sqrt{n}}

0.05\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.05}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}

n = 96.04

Rounding up, 97

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3 years ago
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L = 22 inches

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3 years ago
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