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Step2247 [10]
4 years ago
7

A company is providing gift wrapping services. Box A measures 8 inches by 6.25 inches by 10 inches. Box B measures 9 inches by 5

.5 inches by 11.75 inches. If the company have to wrap 50 boxes and wrapping paper cost $0.03 per square inch. How much money does the company save by choosing to wrap Box A in comparison to Box B?

Mathematics
1 answer:
soldi70 [24.7K]4 years ago
8 0
1. To answer this question you will need to find the surface area of both boxes.

2. You will then take the surface area and multiply it by the number of boxes and then by the cost per square inch for wrapping paper.

3. Finally, subtract the 2 amounts to find out your savings for using Box B.

The answer is that you will save $217.87.

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How do i solve this​
Troyanec [42]

Answer:

f=11

Step-by-step explanation:

If t=7, we can substitute it into the original equation:

f=(2)(7)-3

Simplify:

f=14-3

f=11

<em>~Stay golden~ :)</em>

3 0
3 years ago
50 POINTS PLEASE HELP!!!!!!!!!! ALL SH!+Y ANSWERS REPORTED.
NNADVOKAT [17]
62=14 + 2x

62-14=2x

48=2x

48/2=x

24 answer: 24

hope this helped
6 0
3 years ago
One gallon of water weighs 8.34 pounds. How much does 6 gallons of water weigh? Show your work, including labeling your answer w
maxonik [38]

Answer:

50.04 pounds

Step-by-step explanation:

If 1 gallon weighs 8.34 pounds,

all you have to do is multiply 8.34 by 6

to get what 6 gallons would weigh.

8 0
4 years ago
Data collected over time on the utilization of a computer core (as a proportion of the total capacity) were found to possess a r
Marianna [84]

Answer:

The probability that the proportion of the core being used at any particular time will be less than 0.10 is 0.08146

Step-by-step explanation:

\mu =\frac{\alpha }{\alpha +\beta  } = \frac{1}{1 + \frac{\beta}{\alpha} } 1/3   0.33  = 33.33 %

       The Probability of that the proportion of the core being used at any particular time will be less than 0.10 is given by  

PDF = \frac{x^{\alpha -1} (1-x)^{\beta -1} }{\int\limits^1_0 {u^{\alpha -1} (1-u)^{\beta -1}} \, du }

where x = 0.1

α = 2 and β = 4

PDF = \frac{0.0729 }{\int\limits^1_0 {u^{\alpha -1} (1-u)^{\beta -1}} \, du } = 1.458  

CDF = \frac{\int\limits^{0.1}_0 {t^{\alpha -1} (1-t)^{\beta -1}} \, du }{\int\limits^1_0 {u^{\alpha -1} (1-u)^{\beta -1}} \, du } = 0.08146

The probability that the proportion of the core being used at any particular time will be less than 0.10 = 0.08146.

3 0
4 years ago
Simplify.
Lubov Fominskaja [6]

the answer for this question in 9x-18 (a)

hope it helps you!!!!!!!!!

6 0
3 years ago
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