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3 years ago

What are the factors of 2x + 3x - 54? Select two options

Mathematics

2 answers:

tiny-mole [99]3 years ago

7 0

Answer:

The answer: The factors are (2x-9) and (x+6).

The Problem:

Factor $2x^2+3x-54$

Step-by-step explanation:

So I'm going to do trial factors using the choices to aid me.

Factored form for this problem if it exist will be in the form:

$(mx+n)(kx+p)$ .

In general this is what it would look like if we factored any quadratic in terms of $x$ (given the quadratic is not prime but technically you could factor even over the complex numbers).

Let's look at:

$(mx+n)(kx+p)$

We want to choose $k \text{ and } m$ such that when you multiply them you get 2. Well those would have to be 2 and 1.

$(2x+n)(x+p)$

we want to choose $n \text{ and } p$ such that when you multiply them you get -54. Based on the choices we want to get with -9 and 6, or 9 and -6. We don't know the order we want to choose it in either.

For example which of these would work:

$(2x-6)(x+9)$

$(2x+6)(x-9)$

$(2x-9)(x+6)$

$(2x+9)(x-6)$

We are going to consider only the outer and inner of FOIL since we already know the first times the first is $2x^2$ and the last times the last is $-54$ .

Let's test the first one:

$(2x-6)(x+9)$

Outer: 2x(9)=18x

Inner: -6(x)=-6x

------------------------ADD!

18x-6x=12x

The first choice did not give us the middle term 3x.

Trying the second one would give us the opposite since they are in the same form as previous just the + and - are switched.

Let's look at the third one:

$(2x-9)(x+6)$

Outer: 2x(6)=12x

Inner: -9(x)=-9x

--------------------------ADD!

12x-9x=3x

This is the winner.

The answer: The factors are (2x-9) and (x+6).

sasho [114]3 years ago

4 0

Answer:

a&e

Step-by-step explanation:

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

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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

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