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3 years ago

1/2(-3/5x+1)+7/4=31/12+1/5x. Solve for X.

2 answers:

5 0

Multiply everything by 12 * 5 if you want to get rid of fractions - it's not needed, but i do prefer dealing with integers. Regardless of that previous step, you distribute the first group (1/2 * -3/5x + 1/2 * 1). Then separate x to one side of the equal sign, everything else to the other side.

sdas [7]3 years ago

4 0

For this equation x will equal to x=-2/3

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Carlos earns $11 per hour at his job. Let x represent the number of hours he worked in May. Let y represent the number of hours

Firlakuza [10]

11x+11y

11(x+y)

Explanation

Step 1

Let

Carlos earns == 11 per hour

x represents the number of hours he worked in May

the,

the amount he earned in Mayis

$11\cdot x=11x$

y represent the number of hours he worked in June.

the amount he earned in June was

$11\cdot y=11y$

Step 2

the amount of money he earned is May and June is the sum of the values

$\begin{gathered} \text{total}= \\ 11x+11y \\ \text{if we factorize 11} \\ 11(x+y) \end{gathered}$

I hope this helps you

5 0

1 year ago

You spend 15 minutes doing homework. You spend 3 hours watching television. Write the ratio of the amount of time spent doing ho

german

The ratio of the amount of time spent doing homework to the amount of time spent watching TV is 1/12.

The time spent on doing the homework is 15 minutes.
The time spent watching television is 3 hours.
To calculate the ratio, we must have the same units for both quantities.
We have to convert the time spent watching television from hours to minutes.
1 hour is 60 minutes.
3 hours has 3*60 minutes.
3 hours have 180 minutes.
So, the time spent watching television is 180 minutes.
The ratio of the amount of time spent doing homework to the amount of time spent watching TV is 15/180.
We need to simplify the fraction.
15/180 is equal to 1/12.
Thus, the ratio of the amount of time spent doing homework to the amount of time spent watching TV is 1/12.

To learn more about fractions, visit :

brainly.com/question/10354322

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3 0

1 year 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

Brook makes this conjecture: Because the angles are equal, the proportion of area of each sector relative to the area of the giv

AnnZ [28]

Answer:

The Brooks conjecture is true as the proportion of the area of a sector to the area of a circle is only dependent on the angle.

Step-by-step explanation:

The area of a sector for a radius r and an angle θ is given as

$\Delta Sector=\dfrac{1}{2}r^2\theta$

Whereas the area of a circle for a radius r is given as

$\Delta Circle=\pi r^2$

Now the ratio is given as

$\dfrac{\Delta Sector}{\Delta Circle}=\dfrac{\dfrac{1}{2}r^2\theta}{\pi r^2}$

Here as r is the radius of the circle and it is the same therefore the above ratio is simplified to

$\dfrac{\Delta Sector}{\Delta Circle}=\dfrac{\theta}{2\pi }$

Here as π is constant the only variable here is θ. Thus according to Brooks conjecture when the angles are equal, the proportion of area of each sector relative to the area of the given circle is also equal is true.

6 0

3 years ago

Sklyer has 15 less than twice as many customers as when she started selling newspapers. She now has 155 customers. How many did

arlik [135]

155 = 2x -15
170 = 2x
85 = x

She had 85 when she first started.

8 0

3 years ago

Read 2 more answers