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3 years ago

Solve for x: 5 over x equals 4 over quantity x plus 3 5 3 −3 −15

Mathematics

1 answer:

Firdavs [7]3 years ago

5 0

Answer:

x = - 15

Step-by-step explanation:

The equation is $\frac{5}{x}=\frac{4}{x+3}$

We now cross mulitply and do algebra to figure the value of x (shown below):

$\frac{5}{x}=\frac{4}{x+3}\\5(x+3)=4(x)\\5x+15=4x\\5x-4x=-15\\x=-15$

Hence x = -15

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Which equation shows the point-slope form of the line that passes through (3, 2) and has a slope of 1/3

Vikentia [17]

Answer:

y-2=1/3(x-3)

Step-by-step explanation:

Point slope formula: y-y1=m(x-x1)

m is slope.

y-2=1/3(x-3)

Hope this helps!

If not, I am sorry.

3 0

2 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

3 years ago

Suppose it costs $43 to roll a pair of dice. you get paid $6 times the sum of the numbers that appear on the dice. is it a fair

lutik1710 [3]

The game is not fair as 6 times the expected sum is less than the cost, $43.

In the question, we are asked if the game is fair.

For the game to be fair, 6 times the expected sum for the pair from the game has to be greater than or equal to $43, that is,

6E(X) ≥ 43.

The expected sum for the pair, E(X) can be calculated using the formula,

E(X) = ∑x.p(x),

or, E(X) = 1/18 + 1/6 + 1/3 + 5/9 + 1/6 + 7/9 + 10/9 + 1 + 5/6 + 11/18 + 1/3,

or, E(X) = 107/18.

Now, 6E(X) = 6*(107/18) = 107/3 = 35.67.

Since, the total return from the game is $35.67, which is less than the cost of $43, the game is not fair.

Learn more about expected return from a game at

brainly.com/question/24855677

#SPJ4

7 0

2 years ago

What is the domain of y equals the square root of 4x

pshichka [43]

X>_ 0
If you don’t understand that then x is greater than or equal to 0

4 0

3 years ago

Read 2 more answers

PLEASE ANSWER, FIRST WILL GET A BRAINLIEST, + 2 THANKS (Teaser) An angle that measures 6° is drawn on a piece of paper. You are

Tomtit [17]

Answer: same. ( 6 degree )

Step-by-step explanation:

According to the question, An angle that measures 6° is drawn on a piece of paper. As you are looking at the angle through a×6 magnifying glass, which means it magnifies by a factor of 6 indeed. The length and the width of the shape may magnify by factor of 6, but the degree measure of the angle will remain the same.

Even if it is magnified by a factor of 100, the measure of the angle you see on the paper when you look through the magnifying glass will remain the same.

5 0

4 years ago