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icang [17]
3 years ago
6

What proportion of students are willing to report cheating by other students? A project put this question to an SRS of 180 under

graduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 25 answered, "Yes."
Mathematics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

13.89% of students are willing to report cheating by other students.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 180

Number of students who reported cheating, x = 25

We have to find the proportion of the students are willing to report cheating by other students.

Proportion of students can be calculate as

p = \dfrac{x}{n} = \dfrac{25}{180} = 0.1389\\\\\p = 0.1389\times 100\5\\p = 13.89\%

Thus, 13.89% of students are willing to report cheating by other students.

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faust18 [17]

Answer:

1.65

Step-by-step explanation:

100 pounds of the mixture = 1.36 * 100 = 136 dollars.

She uses 20 pounds of jelly beans that cost

20 * .20   = 4.00

136 - 4 = 132 dollars left for 80 pounds of sugar candies.

80 pounds = 132 pounds of sugar candies.

1 pound =  x

80/1 = 132 / x             Cross multiply

80x = 132

x = 132 / 80

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3 years ago
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damaskus [11]
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Answer:

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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.
Alexandra [31]
37x + 45y = 7940
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Since 45y is in both equations we can more easily transfer from first equation to 2nd equation.

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Simplify by combining 75x - 37x and you have 38x + 7940 = 10,410

Subtract 7940 from both sides and you have 38x = 2470.

Divide both sides by 38 and you have x = 65

To solve for y, take you first equation (or even the second one) and substitute c for 65.

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2 years ago
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Answer:(12,5)

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