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3 years ago

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Eden opened a new checking account on Monday and deposited $25 into it. The table below shows the activity in her account for th

e week. Use this information to determine which of the following statements are true Select two that apply.

1 answer:

exis [7]3 years ago

5 0

If I should solve i need the table given to you

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Consider the two functions. Which statement is true?

DIA [1.3K]

The answer is B i just did it

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3 years ago

Answer the above question

igor_vitrenko [27]

Answer:

B

Step-by-step explanation:

A and C have the primes on the points not the original shape on the point so their wrong

And D the points are marked wrong

4 0

3 years ago

Read 2 more answers

A cone has a volume of 602.88 cubic centimeters and a radius of 6 centimeters. What is its height? Use ≈ 3.14 and round your a

meriva

Answer:

15.98cm

Step-by-step explanation:

Given data

Volume= 602.88cm^3

Radius= 6cm

The formula for the volume of a cone is

V= 1/3 πr^2h

substitute

602.88= 1/3*3.142*6^2*h

602.88=1/3*3.142*36*h

602.88=37.704*h

h= 602.88/37.704

h= 15.98cm

Hence the height is 15.98cm

6 0

3 years ago

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) <u>5.30 years</u>.

(2) (b) <u>0.289</u>.

(3) (b) <u>0.80</u>.

(4) (d) <u>0.50</u>.

(5) (a) <u>5.25 years</u>.

Step-by-step explanation:

Let <em>X</em> = age of the children in kindergarten on the first day of school.

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

The probability density function function of <em>X</em> 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 <em>X</em> 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) <u>5.30 years</u>.

(2)

The standard deviation of a Uniform random variable is:

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

Compute the standard deviation of <em>X</em> 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) <u>0.289</u>.

(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) <u>0.80</u>.

(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) <u>0.50</u>.

(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 <em>x</em> 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) <u>5.25 years</u>.

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4 years ago

Write a algebraic expression for each verbal expression

sashaice [31]

Answer: I don't see the verbal expression

Step-by-step explanation:

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4 years ago