All categories

3 years ago

If you do all o these I will give you 5000 points every month

Mathematics

2 answers:

saw5 [17]3 years ago

6 0

1. find the LCD of 3 and 4 (12). so now your fractions are 5 4/12 and 3 9/12 perimeter is 2l + 2w so 2 (5 4/12) + 2( 3 9/12) = 10 4/12 + 7 6/12 = 17 10/12 or 17 5/6. this is reasonable because that the average perimeter of a sandbox.
2. LCD again is 12, so the fractions are 5 10/12 and 2 1/12. turn those into mixed fractions and get 70/12 and 25/12. 45/12 or 3 3/4. So she babysits 3 3/4 hours during the week.
3. The LCD for this is 8. now the fractions are 2/8 and 1 7/8 or 15/8. add 15/8 and 2/8 to get 17/8 or 2 1/8. He is using a total of 2 1/8 cups of flour.
4. 0.2
5. 0.054
6. 0.024

dont give me those points XD

stealth61 [152]3 years ago

5 0

You might be interested in

in a total of $580,800 last year. If they served an average of 200 customers each day over the 363 days that the deli was open,

brilliants [131]

I think it would be $8.

7 0

3 years ago

Read 2 more answers

How do I solve for x?

Vitek1552 [10]

I guess you multiply everything times "x", the Xs cancel out, then you will have " a + b = 1x " basically the one does not make any change, so you will end with a " a + b = x "

4 0

3 years ago

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

Each pack cost 3. he spent a total of 18 on gum. how many packs of gum did he buy.

TEA [102]

Answer:

So to figure that out you would take the money that he spend which was 18, each pack of gum cost 3, so to figure out the question you would simply divide 18 by 3 which would give you 6.

so how would you check that answer you could take 6 and multiple by the cost of each pack which would give you 18.

so answer is 6.

Step-by-step explanation:

18/3

7 0

3 years ago

What is the area, in square inches, of the trapezoid with th

nekit [7.7K]

Answer:

To find the area of a trapezoid, take the sum of its bases, multiply the sum by ... or area_trapezoid2.gif. Where b1.gif is base1.gif , b2.gif is base2.gif , h.gif ... of a trapezoid with bases of 9 centimeters and 7 centimeters, and a height of 3 ... The area of a trapezoid is

Step-by-step explanation:

3 0

3 years ago