All categories

3 years ago

What does it mean to be a decay type of exponential function?

1 answer:

4 0

Answer:The exponential decay is a model in which the exponential function plays a key role and is one very useful model that fits many real life application theories. The most famous application of exponential decay has to do with the behavior of radioactive materials.

Step-by-step explanation:

You might be interested in

$12 for 6 bagels; $9 for 24 bagels.

BabaBlast [244]

Answer:

What is the question that you are asking?

Step-by-step explanation:

3 0

2 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

Match the following exponential functions with their "b" value.

marishachu [46]

9514 1404 393

Answer:

in order by function: 0.6, 8, 7/3, 2, 1/2

Step-by-step explanation:

The functions are given in the form ...

f(x) = a·b^x

The "b" value is the value immediately to the left of the exponent. (It's not rocket science; it's pattern matching.) Note that the minus sign in g(x) is part of 'a', not part of 'b'.

From the top-down, the functions listed on the left have the b-values shown above.

4 0

2 years ago

For each sequence, find the first 4 terms and the 10th term<br> 1) 2n-1

Katena32 [7]

Answer:

1st term: 1, 2nd term: 3, 3rd term: 5, 4th term: 7 & 10th term: 19

Step-by-step explanation:

1st term: 2(1) - 1

2 - 1 = 1

2nd term: 2(2) - 1

4 - 1 = 3

3rd term: 2(3) - 1

6 - 1 = 5

4th term: 2(4) - 1

8 - 1 = 7

10th term: 2(10) -1

20 - 1 = 19

6 0

2 years ago

A wall of a building is 34 inches wide Sixteen inches is concrete, 12 inches is brick, and 6 inches is limestone What fraction o

Elza [17]

The fraction wall of the concrete is 8/17.

6 0

3 years ago