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kaheart [24]
3 years ago
10

Calculate the volume of each rectangular prism using the information that is provided.

Mathematics
1 answer:
Oksanka [162]3 years ago
5 0

Answer:

a. 224 m³

b. 2366 in³

Step-by-step explanation:

Since,

The volume of a rectangular prism,

V=B\times H

Where,

B = Base area,

H = height

a. Given,

The face area or the base area,

B = 56 square meters,

H = 4 meters,

So, the volume of the rectangular prism,

V=56\times 4=224\text{ cube meters}

b.

B = 169 square inches,

H = 14 inches,

So, the volume of the rectangular prism,

V=169\times 14=2366\text{ cube inches}

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Need answer for this ratios question
Tom [10]

Answer: 4:1

Step-by-step explanation: I think the answer is 4:1. I think this because 20% of 2500, would be 500. So 2500-500=2000. So the ratio would be 2000:500. Then we simplify 2000:500, by dividing 2000 by 500, and dividing 500 by 500. 500 / 500 = 1. 2000 / 500 = 4, so the answer is 4:1. (I am sorry if I am incorrect)

7 0
3 years ago
How many cubes were used to make this figure?
Paha777 [63]

Answer:

I think its 30 I'm not 100% sure tho

3 0
3 years ago
Read 2 more answers
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dem82 [27]
I believe it should be B because it says one more than twice the number so that means addition. Also it says twice the number so that should be 2w.
I’m not 100% sure but it is my best guess
7 0
3 years ago
Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t
Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let \sigma_{1}^{2} be the variance for the population of weight gains for rats given a low dose, and \sigma_{2}^{2} the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2} vs H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}. We have that the sample standard deviation for n_{2} = 22 female control rats was s_{2} = 28 g and for n_{1} = 18 female low-dose rats was s_{1} = 51 g. So, we have observed the value

F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176 which comes from a F distribution with n_{1} - 1 = 18 - 1 = 17 degrees of freedom (numerator) and n_{2} - 1 = 22 - 1 = 21 degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0
3 years ago
The function h(x) = g(x + 2) - 1, complete the table for h(x).
Dafna1 [17]

Answer:

x               g(x + 2)-1               h(x)

-2              g(-2+2)-1              -1

-1               g(-1+2)-1               0

0               g(0+2)-1               1

1                g(1+2)-1                2

2               g(2+2)-1               3

Part A: 3

Part B: -1

Step-by-step explanation:

When you have questions like these all you have to do is plug in and solve.

For example I will do h(2) & h(-2) for reference.

<em><u>h(2):</u></em>

(2+2)-1\\4-1\\3

<em><u>h(-2):</u></em>

(-2+2)-1\\0-1\\-1

4 0
3 years ago
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