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1 year ago

The volume of the shape below is 6 square units. Determine the volume of the shape in two different ways.

Mathematics

1 answer:

REY [17]1 year ago

5 0

The volume of the shape is 6 cubic units

<h3>How to determine the volume of the shape?</h3>

<u>Method 1</u>

From the figure, we can see that:

There are 6 cubes in the figure
The dimensions of the cubes are equal
The volume of each cube is 1 cubic unit

So, the volume of the shape is

Volume = Number of cubes * Volume of each cube

Substitute the known values in the above equation

Volume = 6 * 1

Evaluate

Volume = 6

<u>Method 2</u>

From the figure, we can see that:

There are 6 cubes in the figure
2 cubes at the top and 4 at the bottom
The dimensions of the cubes are equal
The volume of each cube is 1 cubic unit

So, the volume of the shape is

Volume = Top cubes + Bottom cubes

This gives

Volume = 2 * Volume of each cube + 4 * Volume of each cube

Substitute the known values in the above equation

Volume = 2 * 1 + 4 * 1

Evaluate

Volume = 6

Hence, the volume of the shape is 6 cubic units

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Answer:

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Step-by-step explanation:

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Find the area and the circumference of a circle with a diameter 8yrd.

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Answer:

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Step-by-step explanation:

step 1

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substitute

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3 years ago

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3 years ago

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Answer:

Step-by-step explanation:

H₀ : The distribution of the sample agrees with the population distribution.

H₁ : the distribution of the does not agree with the population distribution.

Using the Chi-square test statistics

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Fe = Npi = 1000×51% = 5100

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Fe = Npi = 1000×30% = 300

expected frequency for Idaho

Fe = Npi = 1000×11% = 110

expected frequency for Montana

Fe = Npi = 1000×8% = 80

The Chi=square test statistics is;

x² = ∑ [ (Fo - Fe)² / Fe )

now we substitute

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x² = (450-510)²/510 + (340-300)²/300 + (150-110)²/110 + (60-80)²/80

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first we compute for degree of freedom

d.f = k -1 = 4 - 1 = 3

So from the critical value table, degree of freedom 3 and significance level 0.05,

the chi-square critical value is 7.8147

Therefore chi-square test statistics 31.9376 is greater than the chi-square critical value 7.8147, so NULL HYPOTHESIS IS REJECTED at 5% significance.

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