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3 years ago

11

Consider the function f(x) = 10x^2 + 6x - 2 when x=3

1 answer:

Savatey [412]3 years ago

5 0

Answer:

106

Step-by-step explanation:

Substitute x = 3 into f(x) and evaluate , that is

f(3)

= 10(3)² + 6(3) - 2 = 10(9) + 18 - 2 = 90 + 18 - 2 = 106

You might be interested in

Write the following as a complex number in the form a + bi square root -245

Pepsi [2]

Answer:

Step-by-step explanation:

√(-245)=√(49×5×i^2)=7√5 i=0+7√5 i

5 0

3 years ago

Read 2 more answers

The sum of five consecutive multiples of 4 is 20.Find the smallest integer.

Aloiza [94]

Answer:

The smallest integer is -4.

Step-by-step explanation:

Let the smallest integer be x.

Since the integers are all multiples of 4, they are

=> x, x+4, x+8, x+12 and x+16

=> x + x+4 + x+8 + x+12 + x+16 = 20

=> 5x + 40 = 20

Subtract 40 from both sides of the equation

5x +40 - 40 = 20 - 40

5x = -20

Divide both sides by the coefficient of x(which is 5)

$\frac{5x}{5} = \frac{-20}{5}$

x = -4

∴ the smallest integer is -4.

Hope this helps!!!

6 0

2 years ago

Why does 0.25 x 0.4 have a one decimal place in the product

Dafna11 [192]

The product of these numbers are 0.100
since 0.100 is same as 0.1
so there is only one decimal place
COMMON SENSE

4 0

3 years ago

Which expression are equivalent to the given expression square root of 40

pickupchik [31]

I need picture!!!!!!!!!!

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