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2 years ago

Why does 2(2x+3)=4x+6

Mathematics

2 answers:

PilotLPTM [1.2K]2 years ago

8 0

The 2 outside the brackets means that everything within the brackets needs to be multiplied by 2

marysya [2.9K]2 years ago

3 0

Answer:

Infinite many solutions/ all real numbers/ 6=6

Step-by-step explanation:

Distribute

4x+6=4x+6

-4x -4x

6=6

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Which point is an x-intercept of the quadratic function

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The x-intercept of f(x) = (x + 6)(x - 3)
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3 years ago

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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

What is the solution to this inequality?

solong [7]

Answer: x>−4

Step-by-step explanation:

Let's solve your inequality step-by-step.

6x−5>−29

Step 1: Add 5 to both sides.

6x−5+5>−29+5

6x>−24

Step 2: Divide both sides by 6.

6x/6>-24/6

7 0

3 years ago

Why does 2(2x+3)=4x+6​

Why does 2(2x+3)=4x+6