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2 years ago

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ASAP Please answer. Three students sit in the first row. In the second row, 3 students sit behind each of those students, and so

on. Write an expression with exponents to show how many students sit in the third row? How many total students are seated?

2 answers:

Marizza181 [45]2 years ago

7 0

Answer:

3^x. 39

Step-by-step explanation:

To find the expression, we know the first row is 3 and the next row is 9. So the pattern is multiply by 3. That means the expression is 3^x. Following that, we know the third row is 27. 3 + 9 + 27 = 39

murzikaleks [220]2 years ago

3 0

Answer:

3^2, 9 students.

Step-by-step explanation:

I'm not 100 percent sure about this but I'm pretty sure that because there are 3 rows you can just do 3^2 which is 9.

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I need help with this.

ehidna [41]

Answer:

A and b Are in quadrant 2. F and D are in quadrant 1. F is in quadrant 3 and C is in quadrant 4

Step-by-step explanation:

each quadrant is in the boxes and the question is asking what is each coordinate is in what quadrant

6 0

2 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

When testing whether the correlation coefficient differs from zero, the value of the test statistic is t20=1.95 with a correspon

expeople1 [14]

Given:
value of the test statistic is t20 = 1.95
corresponding p-value of 0.0653 at the 5% significance level

No, I can't conclude that the correlation coefficient differs from zero because the p-value, which is 0.0653 exceeds 0.05.

The correlation coefficient differs from zero when the p-value is less than 0.05.

3 0

2 years ago

Find the amount of tax and the tax rate. Round to two decimal places

Makovka662 [10]

Answer:

The tax amount is 15.17$

The tax rate is 12.4%.

Step-by-step explanation:

First we subtract the cost from the selling price.

137.17 - 122 = 15.17

We have found the tax amount, now for the tax rate.

If tax rate is x, then

122 * x = 137.17

Divide both sides by 122 to get

x = 137.17/122 = 1.124, or .124.

Then convert into percentage by moving the decimal to the right 2 times.

.124 = 12.4%

Hope this helps you

3 0

3 years ago

Read 2 more answers

21(2-y)+12y=44 find y

miss Akunina [59]

Answer: y= -2/9
Explanation:
21(2-y)+12y=44
42-21y+12y=44
42-9y=44
-9y=2
y=-2/9

8 0

2 years ago

Read 2 more answers