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3 years ago

13. Solve the equation: y = 4x + ax + 6 for a.

Mathematics

1 answer:

I am Lyosha [343]3 years ago

4 0

Answer:

a= (y-4x-6)/X

Step-by-step explanation:

Make ax the subject of the formulae

-ax=4x+6-y

Multiply both sides by minus (-)

ax= y-4x-6

Divide both sides by x

a= (y-4x-6)/x

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William got an 85 and an 88 on the first two quizzes. What formula can William use to determine the score he needs on the third

Andrej [43]

He can use Mean Median and Range

8 0

2 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

Can someone explain how to answer this

olya-2409 [2.1K]

Answer:

q= -ck+dp/

-k+d

Step-by-step explanation:

Step 1:

Multiply both sides by p.

dp=ck−kq+pq

Step 2:

Flip the equation.

ck−kq+pq=dp

Step 3:

Add -ck to both sides.

−kq+pq=−ck+dp

Step 4:

Factor out variable q.

q(−k+p)=−ck+dp

Step 5:

Divide both sides by -k+p.

q= -ck+dp/

-k+d

Sorry if it's all letters and u needed numbers.

7 0

2 years ago

The starting salary at McDonald's is $12,800 per year. The company automatically gives you a raise of 1.2%

Sergeeva-Olga [200]

Answer:

paid 8 bucks an hour 69 thousand 14 years

5 0

2 years ago

The value of Desmond's home is modeled by the function V (x) = 275,000(1.09) ²/³x, where x is the number of years since 2012. By

Tasya [4]

Answer:

Step-by-step explanation:

<u>Given function</u>

V(x) = 275000(1.09)^(2/3x)

<u>Every year the value is changing at the rate:</u>

(1.09)^(2/3) = 1.059 ≈ 1.06

The rate of 1.06 is equivalent of 6% increase per year

4 0

2 years ago