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3 years ago

How do you simplify this?????

Mathematics

1 answer:

kap26 [50]3 years ago

5 0

You have to break up 384 into numbers that can be taken out to the radical.

You can break up 384 into 2^3*2^3*6.

Since two 2’s can be taken you would have 4 on the outside and a 6 and x^4 left on the inside.

x^3 can be taken out, leaving an x inside the radical.

The final answer would be 4x on the outside and 6x left under the cubed radical

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It has been found that 85.6% of all enrolled college

NNADVOKAT [17]

Answer:

it has been found that 85.6% of all enrolled college and university students in the U.S. are undergraduates. A random sample of 500 enrolled college students in a particular state revealed that 420 of them were undergraduates. Is there sufficient evidence to conclude that the population differs from the national percentages? Use a = 0.05

6 0

3 years ago

How do I do this problem?

Oduvanchick [21]

$4b+18\leq-12b-14\leq14-5b\\4b+18\leq-12b-14\,and\,-12b-14\leq14-5b\\4b+12b\leq-14-18\,and\,-14-14\leq-5b+12b\\16b\leq-32\,and\,-28\leq7b\\b\leq-\frac{32}{16}\,and\,-\frac{28}{7}\leq b\\b\leq-2\,and\,b\geq-4\\-4\leq b \leq-2$

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3 years ago

which of the following are right triangles: a triangle with side lenghts 6 inches, 8 inches,10 inches, a triangle with side leng

amid [387]

9514 1404 393

Answer:

(6, 8, 10) right
(8, 15, 17) right
(5, 12, 13) right

Step-by-step explanation:

It can be helpful to memorize a few of the Pythagorean triples that show up in Algebra problems. Their multiples also form right triangles.

(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41)

The triple (6, 8, 10) is the first one of these, multiplied by 2 inches.

The only triple that is a sequence of consecutive integers is (3, 4, 5), so you know immediately that (4, 5, 6) does not form a right triangle.

Of course, you can check any of them to see if the squares of the smaller two numbers total the square of the larger number:

4² +5² = 16 +25 = 41 ≠ 6² . . . . . (4, 5, 6) is not a right triangle

5 0

3 years ago

Given that D(x) = 2x, select all of the following that are true statements

Keith_Richards [23]

Answer:

D(x) is a function

D(x) is a direct variation

D(x) is a rule for the set of points (5.10). (6, 12) and (-2,-4)

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

In this problem

$D(x)=2x$

D(x) is a direct variation (linear function)

The variable x is the independent variable or input value

The variable D(x) is the dependent variable or output value

<u><em>Verify each statement</em></u>

case 1) D(x) is a function

The statement is true

see the explanation

case 2) x is the dependent variable

The statement is false

The variable x is the independent variable or input value

case 3) D(x) is a direct variation

The statement is true

see the explanation

case 4) D(x) is a rule for the set of points (5.10). (6, 12) and (-2,-4)

The statement is true

Because

For x=5

substitute in the linear equation

$D(x)=2(5)=10$ ---> is ok

For x=6

substitute in the linear equation

$D(x)=2(6)=12$ ---> is ok

For x=-2

substitute in the linear equation

$D(x)=2(-2)=-4$ ---> is ok

6 0

4 years ago

A random variable X with a probability density function () = {^-x > 0

Sliva [168]

The solutions to the questions are

The probability that X is between 2 and 4 is 0.314
The probability that X exceeds 3 is 0.199
The expected value of X is 2
The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

$P(x) = \int\limits^a_b {f(x) \, dx$

So, we have:

$P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2$

Expand the expression

$P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}$

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

$P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3$

Expand the expression

$P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}$

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

$E(x) = \int\limits^a_b {x * f(x) \, dx$

So, we have:

$E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx$

This gives

$E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx$

Using an integral calculator, we have:

$E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}$

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

$E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx$

This gives

$E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx$

Using an integral calculator, we have:

$E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}$

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

brainly.com/question/15318348

#SPJ1

<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

7 0

2 years ago