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2 years ago

How do I convert 50% to an improper fraction?

Mathematics

1 answer:

densk [106]2 years ago

6 0

50% =
0.5 =
50/100 =
1/2

This cannot be written as an improper fraction

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Which table represents a linear function that has a slope of 5 and a y-intercept of 20?

erastovalidia [21]

Answer:

The first image.

x | y

-4 | 0

0 | 20 <--- When x = 0, the y value is the y-intercept.

4 | 40

8 | 60

Step-by-step explanation:

To check the slope, we can use the equation (y₂ - y₁) ÷ (x₂ - x₁) using any two pairs given. For this example, I'll use (-4, 0) and (4, 40).

x₂ y₂ x₁ y₁

(0 - 40) ÷ (-4 - 4)

(-40) ÷ (-8)

<u>Slope = 5</u>

<u></u>

~Hope this helps!~

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2 years ago

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Which strategy would most help minimize the risk of being underinsured

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Choosing the maximum coverage

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2 years ago

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You discovered you can deduct medical expenses over 6% of your income. Your income is $42,000 and you had medical expenses of $5

iren2701 [21]

No portion of your medical expenses you can deduct

Step-by-step explanation:

You discovered you can deduct medical expenses over 6% of

your income

Your income is $42,000

You had medical expenses of $567

We need to find the portion of your medical expenses you can deduct

∵ Your income = $42,000

∵ You can deduct medical expenses over 6% of your income

- Find the 6% for your income

∵ 6% of the income = $\frac{6}{100}$ × 42,000 = $2520

∴ 6% of your income is $2520

∵ You can deduct medical expenses over 6% of your income

∴ If your medical expenses greater than $2520, then you can

deduct the value greater than it, but if your medical expenses

smaller than $2520, then you can not deduct any value

∵ You had medical expenses of $567

∵ $567 < $2520

∴ There is no value you can deduct from your medical expenses

No portion of your medical expenses you can deduct

Learn more:

You can learn more about percentage in brainly.com/question/12960754

#LearnwithBrainly

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3 years ago

If one letter is chosen at random from the word assists, what is the probability that the letter chosen will be an "s"?

OLEGan [10]

Answer:

4/7 or 57%

There are 7 letters in the word "assists" and 4 of them are "s"

4 divided by 7 will give you the percentage

7 0

3 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

2 years ago