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2 years ago

What is the perimeter of a rectangle in which A= 154 in² and h = 11 in?

Mathematics

1 answer:

Burka [1]2 years ago

5 0

The perimeter of the rectangle is 50 inches

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TRUE OR FALSE GEOMETRY HONORS

nalin [4]

Answer:

I believe it is true

Step-by-step explanation:

A line never stops (extends infinitely in opposite directions)

6 0

3 years ago

Please help! Question above (+10 points)

cricket20 [7]

There are three answers and they are: choice 2, choice 3, choice 5

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Further Explanation:

Choice 1 is false because the intersection of the altitudes of a triangle leads to the orthocenter.

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Choice 2 is true because the three medians of any triangle always intersect at the centroid. A median is a line that goes from one vertex to the midpoint of the opposite side. In this case, we go from point E to the midpoint of side CD. The midpoint of CD is found by bisecting segment CD (see choice 5)

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Choice 3 is true. This is effectively the same as choice 5 below. The "perpendicular" aspect does not matter.

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Choice 4 is false. Following the steps mentioned here will create an altitude line (see choice 1)

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Choice 5 is true. To bisect something is to cut it in half. Let's say that point F is the midpoint of line segment CD. This means that line segment EF is one of the three medians of triangle CDE.

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edit update: I changed choice 3 from false to true

3 0

3 years ago

Consider multiplying 26.2 by 16.43 what would a mathematician say the answer is

IgorC [24]

In mathematics multiplying decimals is just like multiplying a whole number and ignoring the decimals., then just put the decimals in the answer – it is how many decimals does the two numbers combined.

First step:

Line the numbers on the right and do not a line the decimal points

26.2

× 16.43

Second Step:

Multiply the numbers starting on the right, just like multiplying a whole number and ignore the decimals.

26.2

× 16.43

786

1048

1572

+ 262

430466

Third step: add the products

26.2

× 16.43

786

1048

1572

+ 262

430.466 is the mathematician’s answer.

8 0

3 years ago

Round 398.574986215 to the nearest ten-thousand

Alexeev081 [22]

Answer: The answer would be 398.5750

Step-by-step explanation: The ten thousandths place is the fourth digit to the right of the decimal.

7 0

2 years ago

Read 2 more answers

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