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3 years ago

If each face of a cube has in area of 16 square feet, what is the cube's total surface area?

Mathematics

1 answer:

stiv31 [10]3 years ago

4 0

Answer:

96 sq ft

Step-by-step explanation:

There are 6 faces on a cube, and it'd given that each area is 16 sq ft. All you need to do to find total surface area is multiply the two.

6 x 16 = 96

I hope this helped!

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!!!!!!!!!!!!!!!!!!!!!!!!

stira [4]

Answer:

Step-by-step explanation:

it's an acute angle so it has to less than 90 and 71 just seems like it's too much

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2 years ago

14. Two line segments are congruent if they have........... length.

LenaWriter [7]

The same length, if something is congruent it means they’re the same

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2 years ago

Is this right

pychu [463]

Answer:

360 = x+123+ 90

x = 147

Step-by-step explanation:

The three angles form a circle which is 360 degrees

360 = x+123+ 90

Subtract 123 and 90 from each side

360 -123 - 90 = x

147 = x

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3 years ago

Read 2 more answers

78+24-11+56+23-117+1

kakasveta [241]

78+24+11+56+23+117+1 =310

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2 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago