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2 years ago

Select equivalent or nonequivalent for each pair of expressions

Mathematics

1 answer:

wel2 years ago

6 0

Answer:

i think the answer is a as well <3

have a great day!

also wdym by i smell pennies??

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In a 50-gram portion of cereal, 8 grams of it are sugar . Which of the following represents the percent of the 50-grams that is

Mars2501 [29]

Answer:

3) 16

8 is 16% of 50

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3 years ago

Mrs. Kelso and Mr. Bonham gave each their student a small bag of colored tiles. The student each counted the number of purple ti

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Answer:

Step-by-step explanation:

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2 years ago

•Let x represent the number of horses •the number of goats is 5 less than the number of horses •the total number of horses and g

Nesterboy [21]

Answer: x=13

Step-by-step explanation:

(x-5) + x = 21 given

2x-5=21 combined like terms

2x=26 addition property of equality

x=13 division property of equality

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3 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

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3 years ago

Enrollment in a photography class increased from 25 to 35 students. What was the percent of increase in enrollment?

DaniilM [7]

About 29%.
35 - 25 = 10, or the amount of change
10 / 35 = .285, or the amount of change over total

3 0

3 years ago