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Daniel [21]
3 years ago
6

If a backpack was discounted 10% and on sale for $37.80, what is the original price?

Mathematics
1 answer:
Cloud [144]3 years ago
7 0
The original price is @42. To check, find 10% of $42, which is $4.2. Now just subtract $4.2 from the original price, $42, and you get $37.8
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Refer to the number line. Find the coordinate of point Y such that the ratio of MY to YJ is 2:3
balandron [24]

The coordinate of point Y such that the ratio of MY to YJ is 2:3 is 6.4

<h3>How to determine the point?</h3>

The complete question is added as an attachment

The coordinates are given as:

M = 2

J = 18

The ratio is given as:

Ratio = 2 : 3

The location of the point Y is then calculated as:

Y = Ratio * (J - M)

This gives

Y = 2/(2 + 3) * (18 - 2)

Evaluate

Y = 2/5 * 16

This gives

Y = 6.4

Hence, the coordinate of point Y such that the ratio of MY to YJ is 2:3 is 6.4

Read more about line segment ratio at:

brainly.com/question/12959377

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jeka57 [31]
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Please explain to me how to answer this question
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3 years ago
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The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation
Nonamiya [84]

Probabilities are used to determine the chances of events

The given parameters are:

  • Sample size: n = 20
  • Proportion: p = 85%

<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>

Using the binomial probability formula, we have:

P(X = x) = ^nC_x p^x(1 - p)^{n -x}

So, the equation becomes

P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}    

This gives

P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}

P(x = 11) = 0.0011

Express as percentage

P(x = 11) = 0.11\%

Hence, the probability that 11 out of the 20 would graduate is 0.11%

<h3>(b) To what extent do you think the university’s claim is true?</h3>

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

<h3>(c) What is the probability that all  20 would graduate? </h3>

Using the binomial probability formula, we have:

P(X = x) = ^nC_x p^x(1 - p)^{n -x}

So, the equation becomes

P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}    

This gives

P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0

P(x = 20) = 0.0388

Express as percentage

P(x = 20) = 3.88\%

Hence, the probability that all 20 would graduate is 3.88%

<h3>(d) The mean and the standard deviation</h3>

The mean is calculated as:

\mu = np

So, we have:

\mu = 20 \times 85\%

\mu = 17

The standard deviation is calculated as:

\sigma = np(1 - p)

So, we have:

\sigma = 20 \times 85\% \times (1 - 85\%)

\sigma = 20 \times 0.85 \times 0.15

\sigma = 2.55

Hence, the mean and the standard deviation are 17 and 2.55, respectively.

Read more about probabilities at:

brainly.com/question/15246027

8 0
2 years ago
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