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3 years ago

PLEASE HELP ME I WILL GIVE BRAINLEAST !!!!!!!!!!!!1!!!!

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

4 0

Answer:

1 / q^3 = 1 / q^n

n = - 3

Step-by-step explanation:

1/r^-4 = r^4

p^3 * q^2 * p^4 * q^n * r^3 * r^4 = p^7 q ^5 r^7

(p^3 * p^4 *q^2 * r^3 * r^4 ) / ( p^7 * q^5 * r^7) = 1/q^n

(p^7 * q^2 * r^7) / (p^7 * q^5 * r^7) = 1/q^n

q^2 / q^5 = 1/q^n

1 / q^3 = 1 / q^n

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What is the value of 4(x - 2)(x + 3) + 7(x - 2)(x - 5) - 6(x - 3)(x - 5) when x = 5?

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Answer:

Step-by-step explanation

4(x - 2)(x + 3) + 7(x - 2)(x - 5) - 6(x - 3)(x - 5)

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7(5 - 2)(5 - 5) = 0

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What's the output of a system modeled by the function ƒ(x) = x5 – x4 + 9 at x = 2?

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3 years ago

A school has an enrollment of 600 students. 330 of the students are girls. Express the fraction of students who are boys to tota

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3 years ago

Read 2 more answers

The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir

Kazeer [188]

Answer:

(1) (c) 5.30 years.

(2) (b) 0.289.

(3) (b) 0.80.

(4) (d) 0.50.

(5) (a) 5.25 years.

Step-by-step explanation:

Let X = age of the children in kindergarten on the first day of school.

The random variable X follows a continuous Uniform distribution with parameters a = 4.8 years and b = 5.8 years.

The probability density function function of X is:

$f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a$

(1)

The expected value of a Uniform random variable is:

$E(X)=\frac{1}{2}(a+b)$

Compute the mean of X as follows:

$E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3$

Thus, the mean of the distribution is (c) 5.30 years.

(2)

The standard deviation of a Uniform random variable is:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289$

Thus, the standard deviation of the distribution is (b) 0.289.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

$P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\$

$=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8$

Thus, the probability that a randomly selected child is older than 5 years old is (b) 0.80.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

$P(5.2$

$=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5$

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) 0.50.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of x as follows:

$P (X < x) = 0.45$

$\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45$

$\int\limits^{x}_{4.8} {1}\, dx=0.45$

$[x]^{x}_{4.8}=0.45$

$x-4.8=0.45\\$

$x=0.45+4.8\\x=5.25$

Thus, the age of the child at the 45th percentile is (a) 5.25 years.

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3 years ago

Simplify the Matrix Operation 4[-4 5 -6]

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Answer:

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