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3 years ago

The rectangular prism shown has a surface area of 488 square inches. What is the length of the prism?

Mathematics

1 answer:

BigorU [14]3 years ago

8 0

Answer:

length of the prism = 244 - (width x height) / width + height)

Step-by-step explanation:

Let length be a and width be b and height be c

2 (a * b) + 2 (b* c) + 2 (a * c) = 488

2ab+ 2bc + 2ac = 488

ab + bc + ac = 244

ab + ac = 244 - ac

a (b + c) = 244-ac

a = 244-bc / b + c

Therefore the length of the prism = 244 - (width x height) / width + height)

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3 years ago

What is 0.47058824 rounded to nearest tenth

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Answer:

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3 years ago

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REPOST

melamori03 [73]

Answer:

5/48, 3/16, .5, .75, 13
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Step-by-step explanation:

5/48 = 1.0291666666 | 3/16 = .1875 the rest is obvious
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2 years ago

What is the perimeter of a polygon with vertices at (−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1) ?

alekssr [168]

First we need the distances of the sides of the polygon, because perimeter = sum of all sides.
$d = \sqrt{{(y2 - y1)}^{2} + {(x2 - x1)}^{2} }$
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$d3 \: = \sqrt{{(7 - 4)}^{2} + {(2 - 6)}^{2} } \\ = \sqrt{{(3)}^{2} + {( - 4)}^{2} } = \sqrt{9 + 16} \\ = \sqrt{25} \: = 5$
$d4 = \sqrt{{(4 - 1)}^{2} + {(6 - 6)}^{2} } \\ = \sqrt{ {(3)}^{2} + {(0)}^{2} } = \sqrt{9} \: = 3$
$d5 = \sqrt{{(1 - 1)}^{2} + {(-2 - 6)}^{2} } \\ = \sqrt{ {(0)}^{2} + {( - 4)}^{2}} = \sqrt{16} \: = 4$
Now, we add all sides for the perimeter:
p = d1 + d2 + d3 + d4 + d5
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3 years ago

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A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn

Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, $\overline x_1$ = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, $\overline x_2$ = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

$t_{\alpha /2}$ = $t_{0.005, \, 21}$ = 1.721

Therefore, we have;

$\left (6.59- 15.5 \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}$

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey had lower average score than the subjects in the control group

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2 years ago