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2 years ago

Helpp! Evaluate the expression and enter your answer below. 42-13

Mathematics

1 answer:

Andrews [41]2 years ago

5 0

Answer - 3

In order to do this equation you would follow PEMDAS. Being as there isn’t any parentheses, multiplication, or division involved, we only need step E ( exponents) and step S ( subtraction). In the acronym, E comes before S so we multiple 4 * 4 ( if you did not know: the exponent of 2 means you would multiply the number with itself.) 4 * 4 equals 16. The equation is now 16- 13. When you subtract the two, the product is 3.

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Describe the symmetry of these functions

Oduvanchick [21]

If it was line symmetry it would need to repeat the same shape twice, which the first follows but the second doesn’t.
if it was rotational, you would need to be able to take the shape in the top right and rotate it counterclockwise or clockwise to get the shape that locks in place. that doesn’t follow that.
both line and rotational symmetry is incorrect because the first example would need to lock up inside the right side of that example.

the answer is C

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3 years ago

Read 2 more answers

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3 years ago

An online health and medicine company claims that about 50% of Internet users would go to an online sight first for

Serggg [28]

Answer:

The claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

Step-by-step explanation:

Since an online health and medicine company claims that about 50% of Internet users would go to an online sight first for information regarding health and medicine, and a random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet, to test the claim that less than 50% of Internet users get their health questions answered online, the following calculation must be performed:

1318 = 100

606 = X

606 x 100/1318 = X

60600/1318 = X

45.978 = X

Therefore, the claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

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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

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

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3 years ago

Will mark you brainliest

kumpel [21]

Answer: 37%

Step-by-step explanation:

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2 years ago