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3 years ago

What is the answer? (Number 8)

Mathematics

2 answers:

barxatty [35]3 years ago

8 0

A mode is the most often occurring number, so the answer is 14

mixas84 [53]3 years ago

7 0

Hi , the answer is 14 because number 14 the most abundant number , and basically mode is the number that is repeating , so your answer is 14.

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Speed=60 km/h time= 2 hours 20 minutes, what is the distance?

daser333 [38]

Answer:

26km

Step-by-step explanation:

d=speed × time

d=60×2.33

d= 140km

8 0

3 years ago

How to model 4÷2/3??

stealth61 [152]

Answer:

We have 4 x ( 3 / 2 ) = 12 / 2 = 6;

Step-by-step explanation:

4 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

3 years ago

The graph of function g is a vertical stretch of the graph of function f by a factor of 3. Which equation describes function g

anyanavicka [17]

Answer:

Step-by-step explanation:

A vertical stretch by a factor of k means that the point (x, y) on the graph of f (x) is transformed to the point (x, ky) on the graph of g(x), where k <1.

If k =1, then same graph we obtain and if k>1 we get a vertically shrink graph.

In our question, there is a vertical stretch of 3. This means new graph would have points as (x,y/3)

i.e. instead of f(x) = y, we have now f(x) = y/3

So transformation is g(x) = 3f(x)

Option A is correct.

2) Here f(x) = 3x+8 is transformed to g(x) =3x+6

i.e. y intercept is reduced by 2 units. Hence there is a translation of 2 units down.

8 0

3 years ago

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Where are the minimum and maximum values for f(x) = -2 + 4 cos x on the interval (0,21]?

Dafna11 [192]

Answer:

Step-by-step explanation:

Maximum value is when cos x = 1

So it is -2 + 4(1) = 2.

Minimum value, when cos x = -1:

= -2 + 4(-1) = -6.

4 0

3 years ago

Read 2 more answers