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rosijanka [135]

3 years ago

5

WILL MARK BRAINLIEST If you knew the exact values of the data points for this data set, which measure of center (the mean or med

ian) would most likely provide a more accurate picture of the data set? Explain your reasoning. asking for a friend lol

1 answer:

Alexxx [7]3 years ago

8 0

Answer:

Mean

Step-by-step explanation:

If you knew the exact values the mean would provide a more accurate picture because you will have the average of the range of the numbers not the middle of your data points that you would get from finding the median

Hope this helps!

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Find k if (x+1) 2x^3+kx^2+1

Viktor [21]

<h2>Question:</h2>

Find k if (x+1) is a factor of 2x³ + kx² + 1

<h2>Answer:</h2>

k = 1

<h2>Step-by-step explanation:</h2>

The factor of a polynomial F(x) is another polynomial that divides evenly into F(x). For example, x + 3 is a factor of the polynomial x² - 9.

<em>This is because;</em>

i. x² - 9 can be written as (x - 3)(x + 3) which shows that both (x - 3) and (x + 3) are factors.

ii. If x = -3 is substituted into the polynomial x² - 9, the result gives zero. i.e

=> (-3)² - 9

=> (9) - 9 = 0

Therefore, if (x + a) is a factor of a polynomial, substituting x = -a into the polynomial should result to zero. This also means that, if x - a is a factor of a polynomial, substituting x = a into the polynomial should give zero.

<em><u>From the question</u></em>

Given polynomial: 2x³ + kx² + 1

Given factor: x + 1.

Since x + 1 is a factor of the polynomial, substituting x = -1 into the polynomial should give zero and from there we can calculate the value of k. i.e

2(-1)³ + k(-1)² + 1 = 0

2(-1) + k(1) + 1 = 0

-2 + k + 1 = 0

k - 1 = 0

k = 1

Therefore the value of k is 1.

3 0

2 years ago

Ab and ba name the same ray.<br><br><br> always<br> sometimes<br> never

vekshin1

The <em><u>correct answer</u></em> is:

never

Explanation:

A ray is named using the endpoint first and then another point on the ray. This means that ray AB has endpoint A and point B on the ray; ray BA has endpoint B and point A on the ray. These are not the same ray, as they do not have the same endpoint.

7 0

3 years ago

Read 2 more answers

NEED HELP ASAP. By what percent will a fraction change if its numerator is decreased by 30% and its denominator is decreased by

joja [24]

Answer:

0.7x/0.8y = x/y * 0.7/0.8 = x/y * 0.875

So the fraction is reduced by 12.5%

5 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

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

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2 years ago

5 wholes 2/3 gallon in 2 hours

Firdavs [7]

What do you mean? I don't think it's enough information

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3 years ago