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3 years ago

Which could be the dimensions of a rectangular prism to service area is less than 160 ft.² check all that apply

Mathematics

2 answers:

Dovator [93]3 years ago

6 0

Area of a rectangular prism: 2(wl+hl+hw), given w=width, l=length, and h=height.

Plug the answer choices into this formula to find which one(s) are less than 160 sq. feet.

A. 136 sq. feet

B. 188 sq. feet

C. 202 sq. feet

D. 162 sq. feet

E. 142 sq. feet

So, the only ones that are less than 160 sq. feet are A (8 feet by 4 feet by 3 feet) and E (3 feet by 5 feet by 7 feet).

Your answers are A and E.

I hope this helps :)

raketka [301]3 years ago

3 0

Area of rectangular prism is given by:

A=2(wl+hl+lw)

where w=width h=height, l=length

checking for given options

1. 8 ft, by 4ft by 3ft

Area = 136 square feet

5. 3 feet by 5 feet by 7 feet

Area =142 square feet

so our first and last option matches the requirement of area less than 160 square feet.

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How to solve -2(1-7n)-5n=34

OLEGan [10]

-2(1-7n)-5n=34

Multiply the bracket by -2

(-2)(1)=-2

(-2)(-7n)=14n

-2+14n-5n=34

-2+9n=34

Move -2 to the other side. Sign changes from -2 to +2.

-2+2+9n=34+2

9n=34+2

9n=36

Divide by 9 for both sides

9n/9=36/9

n=4

Answer: n=4

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3 years ago

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A bowling ball has a diameter of 8.5 inches. If it is rolled down a 60 foot bowling lane, how many complete revolutions will it

Lemur [1.5K]

Answer:

Step-by-step explanation:

12 in = 1 ft

first you need to convert 60ft to in:

60 * 12 = 720in

then divide 720(the amount of space you are covering by 8.5

720/8.5 = 84.7

then i guess you can round it to a full revolution so it would be 85

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3 years ago

Suppose ABC added a new point, D. If the coordinates of D were (5, 2), what would be the coordinates of D' have to be if it foll

atroni [7]

Answer:

C the coordinates are just fliped since It was flipped on the y- axis

Step-by-step explanation:

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2 years ago

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The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago

What is the answer to this problem?

Andrei [34K]

i think the best answer for this question is A)

5 0

3 years ago