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3 years ago

A rectangle prism has dimensions of 1/2, 2 and 7/2 inches. How many cubes with side edges of 1 inch will fit into the prism?

2 answers:

8 0

The volume of a rectangular prism is the product of its three dimensions.
We have then:
V = (1/2) * (2) * (7/2)
V = 3.5 in ^ 3
On the other hand the volume of a cube is:
V = L ^ 3
Where
L = length of its sides.
V = (1) ^ 3
V = 1 in ^ 3
The number of cubes inside the prism is:
N = (3.5) / (1)
N = 3.5
That is, only 3 cubes of volume 1 in ^ 3 fit inside the prism.
Answer:
3 cubes of volume 1 in ^ 3 inside the prism.

uysha [10]3 years ago

5 0

Dimensions of the rectangular prism: a=1/2 inches, b=2 inches, c=7/2 inchesVolumen of the rectangular prism: Vr=abcVr=(1/2 inches)(2 inches)(7/2 inches)Vr=[ (1*2*7) / (2*2) ] inches^3Vr=14/4 inches^3Vr=7/2 inches^3
Side of the cube: s=1 incheVolume of the cube: Vc=s^3Vc=(1 inche)^3Vc=1 inche^3

Number of cubes will fit into the rectangular prism: nn=Vr/Vcn=(7/2 inches^3) / (1 inche^3)n=7/2n=3.5
3.5 cubes with side lengths of 1 inche will fit into the prism
Read more on Brainly.com - brainly.com/question/9194776#readmore

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If sales are $425,000, variable costs are 62% of sales, and operating income is $50,000, what is the contribution margin ratio?

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In an internal operating income statement, the form is as such:

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Gnoma [55]

Answer:

Step-by-step explanation:

Hello!

The objective is to study whether there is a greater force after impacting on one- handed backhand drive in advanced tennis players than in intermediate tennis players.

Sample 1: Advanced tennis players

X₁: Force (N) on the hand just after impact on a one- handed backhand drive for an advanced tennis player.

n₁= 6

X[bar]₁= 40.29 N

S₁= 11.29

Sample 2: Intermediate players

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Assuming that both variables have a normal distribution and both population variances are equal, to compare these two populations is best to do so trough their population means using a t-test for independent samples.

If the force is greater for the advanced players than for the intermediate players, then you'd expect the population mean for the advanced players to be greater than the population mean for the intermediate players:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.05

$t= \frac{(X_[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{5*127.51+7*68.92}{6+8-2} }= 9.66$

$t_{H_0}= \frac{(40.29-21.40)-0}{9.66\sqrt{\frac{1}{6} +\frac{1}{8} } } = 3.62$

Using the p-value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value for this test is 0.00024, it is less than the level of significance, so the decision is to reject the null hypothesis.

This means that at a 5% significance level you can conclude that the average force experienced on the hand after a one-handed backhand drive for advanced players is greater than the average force experienced on the hand after a one-handed backhand drive for intermediate players.

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