All categories

2 years ago

Greg is 3 years older than Kevin. In seven years, the sum of their ages will be 41. How old is each now ?

Mathematics

1 answer:

Artist 52 [7]2 years ago

3 0

Let their ages be g and k. Then g = k + 3.

Next: (g+7) + (k+7) = 41.

Substituting k + 3 for g in the above equation, we get k + 3 + 7 + k + 7 = 41.

Combining like terms: 2k + 17 = 41. Then 2k = 24, and k = 12.

Kevin is 12 years old and Greg is 15 years.

You might be interested in

A random variable X has a gamma density function with parameters α= 8 and β = 2.

DerKrebs [107]

I know you said "without making any assumptions," but this one is pretty important. Assuming you mean $\alpha,\beta$ are shape/rate parameters (as opposed to shape/scale), the PDF of $X$ is

$f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}$

if $x>0$ , and 0 otherwise.

The MGF of $X$ is given by

$\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx$

Note that the integral converges only when $t$ .

Define

$I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx$

Integrate by parts, with

$u = x^n \implies du = nx^{n-1} \, dx$

$dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}$

so that

$\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}$

Note that

$I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}$

By substitution, we have

$I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}$

and so on, down to

$I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}$

The integral of interest then evaluates to

$\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}$

so the MGF is

$\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}$

The first moment/expectation is given by the first derivative of $M_X(t)$ at $t=0$ .

$\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}$

Variance is defined by

$\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2$

The second moment is given by the second derivative of the MGF at $t=0$ .

$\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18$

Then the variance is

$\Bbb V[X] = 18 - 4^2 = \boxed{2}$

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

$M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k$

where $M_X^{(k)}(0)$ is the $k$ -derivative of the MGF evaluated at $t=0$ . This is also the $k$ -th moment of $X$ .

Recall that for $|t|$ ,

$\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k$

By differentiating both sides 7 times, we get

$\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k$

Then the $k$ -th moment of $X$ is

$M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}$

and we obtain the same results as before,

$\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4$

$\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18$

and the same variance follows.

6 0

2 years ago

Write a fraction to a decimal

Alona [7]

1/2 = 0.5

1/4 = 0.25

1/5 = 0.2

5 0

2 years ago

How much greater is the median number of pairs of shoes in the inventory at the Golf Pro Shop than the median number of pairs of

larisa86 [58]

The median if golf pro shop is 20 pairs greater than tennis pro shop

3 0

2 years ago

There are a total of 375 flamingos, parrots, herons, and chachalacas at the Famous Bird Sanctuary. The ratio of flamingos to par

Tems11 [23]

Answer:

Step-by-step explanation:

Since the ratio between flamingos and chachalacas is 2:1, we see that the ratio of flamingos to chachalacas to parrots to herons is 10:5:7:3. The sum of these numbers is 25, so every 'ratio number' gets 375/25 = 15 animals. So the number of chachalacas is 15 * 5 = 75.

3 0

2 years ago

what is the volume of a cylinder, in cubic inches, with a height of 2 inches and a base diameter of 4 inches? Round to the neare

seraphim [82]

Answer:

V≈25.13in³

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers