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kvasek [131]
3 years ago
6

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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ANEK [815]

Answer:

4..?

Step-by-step explanation:

4 0
3 years ago
What is 0.47058824 rounded to nearest tenth
Butoxors [25]

Answer:

0.5

Step-by-step explanation:

0.47058824

the 4 is in the tenths place

the 7 makes it turn to a 5

7 0
3 years ago
Read 2 more answers
REPOST
melamori03 [73]

Answer:

  1. 5/48, 3/16, .5, .75, 13
  2. 1/5, .35, 12/25,  .5, 4/5
  3. -3/4, -7/10, 3/40, 8/10
  4. -.65, -3/8, 5/16, 2/4

Step-by-step explanation:

  1. 5/48 = 1.0291666666 | 3/16 = .1875 the rest is obvious
  2. 1/5 = .2 | 12/25 = .48 | 4/5 = .8
  3. -3/4 = -.75 | -7/10 = -.7 | 3/40 = .075 | 8/10 = .8
  4. -3/8 = -.375 | 5/16 = .3125 | 2/4 = .5
5 0
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} }
d1 =  \sqrt{{(1 - 4)}^{2} + {( - 2 -  - 2)}^{2} } \\  =  \sqrt{{(- 3)}^{2} + {(0)}^{2} } =  \sqrt{9}  \: = 3
d2 = \sqrt{{(4 - 7)}^{2} + {( - 2 - 2)}^{2} } \\  = \sqrt{{(- 3)}^{2} + {( -4)}^{2} } =  \sqrt{9 + 16}  \\  =  \sqrt{25}  \: = 5
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
p = 3+5+5+3+4 = 20 units





5 0
3 years ago
Read 2 more answers
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

2. The possible issues in the calculations are;

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

b. The sample size is small

5 0
2 years ago
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