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kodGreya [7K]
3 years ago
14

Find the volume of the rectangular prism with a base area of 63 and a height of 14 feet.

Mathematics
1 answer:
gayaneshka [121]3 years ago
5 0

Answer:

63 = 14

devide 14 on both sides

14/14 cancels out

63/14 = 4.5

the answer is 4.5

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Please help it’s easy, State the property that is used. 10=10•1
Molodets [167]

The property identity of multiplication is applied in the given statement.

<h3>Properties of Multiplication</h3>

The properties of multiplication are:

  • Distributive:  a(b±c)=  ab±ac
  • Comutative:   a . b = b. a
  • Associative:    a(b+c)=  c(a+b)
  • Identity: b.1=b

From the property identity, you know that the product between any number by the number 1 equals that number. Example: 3 • 1=3.

The question shows 10=10•1. Like, it was shown previously, this occurs due to the identity property of multiplication.

Read more about the identity property here:

brainly.com/question/23977324

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5 0
2 years ago
Which out of the options is correct? Merci!
AysviL [449]

Answer:

C

Step-by-step explanation:

The answer is C because 1 * 2 = 2, 2 * 2 = 4, and 4 * 2 = 8.

8 0
3 years ago
Of the following points, name all that lie on the same horizontal line? (1,-2) (-1,2) (-2,1) (2,-1)
11Alexandr11 [23.1K]

Answer:

None

Step-by-step explanation:

Let’s graph the points first.

Look at the image below.

And by looking at the image we can tell that none of the points lay on the same horizontal line.

<em>Thus,</em>

<em>none of the points lay on the same horizontal line.</em>

<em />

<em>Hope this helps :)</em>

<em />

8 0
3 years ago
Anna drives 3.9 miles to work. Tanya drives 7.5 times farther to work than Anna drives.
nalin [4]

Answer:Tanya drives 11.4 miles farther.

Step-by-step explanation:

3.9+7.5=11.4 miles

4 0
3 years ago
Read 2 more answers
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
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