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3 years ago

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Consider the following pair of equations: y = −2x 8 y = x − 1 Explain how you will solve the pair of equations by substitution.

Show all the steps and write the solution in (x, y) form.

2 answers:

NARA [144]3 years ago

8 0

If so y= -2x and 8y=x-1...

solution:

$y=-2x \\ \\ 8y=x-1 \Rightarrow \\ \\ 8.(-2x)=x-1 \Rightarrow \\ \\ -16x=x-1 \Rightarrow \\ \\ 17x=1 \Rightarrow \\ \\ x=\dfrac{1}{17}$

givi [52]3 years ago

5 0

Step-by-step explanation:

Consider the following pair of equations: y = −2x 8 y = x − 1 Explain how you will solve the pair of equations by substitution. Show all the steps and write the solution in (x, y) form.

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Polly decides to cut the extra-large pizza into 12 equal slices, how much of the pizza is 2 slices in decimal form?

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Answer: 0.5

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3 years ago

Solve y = x + 8 for x.<br> x = y + 8<br> x = y - 8<br> X = -y + 8<br> X = -V +8

JulijaS [17]

Answer:

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3 years ago

find the area of a parallelogram if a base and corresponding altitude have the indicated lengths base 1 1/2 feet , altitude 6 in

Ostrovityanka [42]

The area of parallelogram is 108 square inches

<em><u>Solution:</u></em>

<em><u>The formula for area of parallelogram is:</u></em>

$Area = base \times height$

From given,

$Base = 1\frac{1}{2}\ feet = \frac{2 \times 1 + 1}{2} = \frac{3}{2}\ feet$

$Height = 6\ inches$

<em><u>Convert feet to inches</u></em>

1 feet = 12 inches

$\frac{3}{2}\ feet = \frac{3}{2} \times 12\ inches = 18\ inches$

<em><u>Therefore, area of parallelogram is:</u></em>

$Area = 18 \times 6\\\\Area = 108$

Thus area of parallelogram is 108 square inches

4 0

3 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

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<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 equation of the translated function, g(x), if

ICE Princess25 [194]

Answer:

D

Step-by-step explanation:

g(x) = (x+4)^2 +6

3 0

3 years ago