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Hoochie [10]
3 years ago
13

PLEASE HELP!!!

Mathematics
1 answer:
allsm [11]3 years ago
4 0

Answer/Step-by-step explanation:

Part A: Net A is the correct net. If we decide to fold the net, we'd get the shape of the prism. Folding back net B won't give us the shape of the prism because of the position and arrangement of the right triangular  bases.

Part B:

AB =  3in

BC = 5 in

CD = 9.4 in

Part C: surface area of the prism can be calculated by calculating the area of each part of the net, and summing them together as follows,

Area of the 2 triangular bases = 2(½*base*height of triangle) = 2(½*4*3) = 2*2*3 = 12 in²

Area of the three rectangles =

= (9.4*4) + (9.4*3) + (9.4*5) = 37.6 + 28.2 + 47 = 112.8 in^2

Surface area of prism = 12 + 112.8 = 124.8 in²

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inessss [21]
The consult must make at least 600 dollars. That means the combination of reviews and new written pieces must be at least $600, or greater than or equal to (<span>≥) $600.

You are also told that each new piece = $120 and each review = $60. Let's use variables n = new piece and r = review. (You might be given different variables for your answer choices). That means the total money made from a combination of new pieces and reviews is:
120n (money made from each new piece times n number of new pieces) + 60r (money made from each review times r number of reviews).

Put that all together to get your inequality:
120n + 60r </span>≥ 600

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Answer: 120n + 60r ≥ 600
8 0
3 years ago
a teacher promised a movie day to the class that did better, on average, on their test. The box plot shows the results of the te
natka813 [3]

The question is incomplete. The complete question is

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.

Which class should get the reward, and why?

- The 2nd-period class should get the reward. They have the highest score, a perfect 100.

- The 2nd-period class should get the reward. They have a higher median.

- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.

The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is

\frac{77+78+89+92+100}{5}=87.2

In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.

Hence, the mean of the results of the 4th period is

\frac{72+83+89+96+98}{5}=87.6

Hence, obviously, the 4th period is having more mean.

Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.

So, just by observing the box plot, we can give the statement that  "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "

Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

Learn more about box plots here-

brainly.com/question/1523909

#SPJ10

6 0
2 years ago
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Veronika [31]

Answer:

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First, you add 35 + 14 + 18 + 27 which is 94.
Next, you divide 94 by the number of deliveries there were. Which is 4.
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I Hope This Helped :D

5 0
2 years ago
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Ivahew [28]
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