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ser-zykov [4K]
3 years ago
5

What is the result when the number 20 is decreased by 75%?

Mathematics
2 answers:
Katen [24]3 years ago
7 0

Answer:

5

Step-by-step explanation:

20 decreased by 75%

20 - 20*75%

Changing to decimal form

20 -20*.75

20 -15

5

The result is 5

fiasKO [112]3 years ago
7 0

Answer:

5

Step-by-step explanation:

100 decreased by 75% is 25% or a quarter so fourths so find a fourth of 20 wich is 5 because 20÷4=5

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173,482-5.5=173,476.5

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You bought a large cooler that holds 55 gallons. How many quarts is that?
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  • a) 200 yards

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2 years ago
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The total length of pencils A, B and C is 29 cm. Pencil a is 11 cm shorter then pencil B, and pencil B is twice as long a pencil
satela [25.4K]

The length of pencil A is 5 cm

<em><u>Solution:</u></em>

Let the length of pencil A be "x"

Let the length of pencil B be "y"

Let the length of pencil C be "z"

<em><u>The total length of pencils A, B and C is 29 cm</u></em>

Therefore,

length of pencil A + length of pencil B + length of pencil C = 29

x + y + z = 29 ------------ eqn 1

<em><u>Pencil A is 11 cm shorter then pencil B</u></em>

x = y - 11 ------- eqn 2

<em><u>Pencil B is twice as long a pencil C</u></em>

y = 2z

z = \frac{y}{2} ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

y - 11 + y + \frac{y}{2} = 29\\\\2y + \frac{y}{2} = 29 + 11\\\\\frac{4y+y}{2} = 40\\\\5y = 80\\\\y = 16

<em><u>Substitute y = 16 in eqn 2</u></em>

x = 16 - 11

x = 5

Thus length of pencil A is 5 cm

5 0
3 years ago
. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
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