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Diano4ka-milaya [45]

2 years ago

6

4444444444444444444444444444444444444444444444444444444444

1 answer:

Liula [17]2 years ago

4 0

If you want to solve that put it in the formula to find the slope
M=
Y1-y2
---------
X1-x2

And plug that into slope int form with one of the points to find the y int/ b value and then your have ur equ

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Show the factors of 5 x y square + 7 x square y by tree diagram

zlopas [31]

Answer:

Answer:

Domain = {x /x ≥ -7}

Step-by-step explanation:

The given function y =

Inside the square, negative will not come.

First let's find the domain. x + 7 must be greater than or equal to 0.

x + 7 ≥ 0

Solving x, we get

x ≥ -7

Domain = {x /x ≥ -7}

Hope this will help you.

3 0

3 years ago

One pump can empty a pool in 4 days, whereas a second pump can empty the pool in 10 days. How long will it take the two pumps, w

gtnhenbr [62]

Answer:

20/7 days (just less than 3 days)

Step-by-step explanation:

Recall that (1 job) = (rate)(time), so time = (1 job) / (rate).

Set up and solve the following equation:

1 job

------------------------------- = time required for 2 pumps working together

1 job 1 job

---------- + -------------

4 days 10 days

This comes out to:

1 job

------------------------------------------- = time required

10 job-days 4 job-days

------------------ + -----------------

40 days 40 days

or:

1 job

-------------------- = (40/14) days, or 20/7 days (just less than 3 days)

14 job·days

-----------------

40 days

7 0

3 years ago

8) The perimeter of a rectangle is 20x2 + xy - 7y2 and one of it's sides is

Sedbober [7]

Answer:

3x^2 + 3xy/2 - 7xy^2/2

Step-by-step explanation:

So we know the perimeter is 20x^2 + xy - 7y^2,

To find any perimeter you need 2l + 2w = P so,

One of the sides is 7x^2 - xy

First plug in the values,

2(7x^2-xy) + 2w = 20x^2 + xy - 7y^2

Multiply,

14x^2-2xy + 2w = 20x^2 + xy - 7y^2

Subtract,

14x^2 - 2xy - 14x^2 + 2xy + 2w = 20x^2 + xy - 7y^2 - 14x^2 + 2xy

2w = 6x^2 + 3xy - 7y^2

w = 3x^2 + 3xy/2 - 7xy^2/2

5 0

3 years ago

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, 5% of the slabs are cla

AnnyKZ [126]

Answer:

(a) Name: Multinomial distribution

Parameters: $p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

(b) Range: $\{(x,y,z)| x + y + z=20\}$

(c) Name: Binomial distribution

Parameters: $p_1 = 5\%$ $n = 20$

$(d)\ E(x) = 1$ $Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1) = 0.7359$

$(h)\ E(Y) = 17$

Step-by-step explanation:

Given

$p_1 = 5\%$

$p_2 = 85\%$

$p_3 = 10\%$

$n = 20$

$X \to$ High Slabs

$Y \to$ Medium Slabs

$Z \to$ Low Slabs

Solving (a): Names and values of joint pdf of X, Y and Z

Given that:

$X \to$ Number of voids considered as high slabs

$Y \to$ Number of voids considered as medium slabs

$Z \to$ Number of voids considered as low slabs

Since the variables are more than 2 (2 means binomial), then the name is multinomial distribution

The parameters are:

$p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

And the mass function is:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

Solving (b): The range of the joint pdf of X, Y and Z

Given that:

$n = 20$

The number of voids (x, y and z) cannot be negative and they must be integers; So:

$x + y + z = n$

$x + y + z = 20$

Hence, the range is:

$\{(x,y,z)| x + y + z=20\}$

Solving (c): Names and values of marginal pdf of X

We have the following parameters attributed to X:

$p_1 = 5\%$ and $n = 20$

Hence, the name is: Binomial distribution

Solving (d): E(x) and Var(x)

In (c), we have:

$p_1 = 5\%$ and $n = 20$

$E(x) = p_1* n$

$E(x) = 5\% * 20$

$E(x) = 1$

$Var(x) = E(x) * (1 - p_1)$

$Var(x) = 1 * (1 - 5\%)$

$Var(x) = 1 * 0.95$

$Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3)$

In (b), we have: $x + y + z = 20$

However, the given values of x in this question implies that:

$x + y + z = 1 + 17 + 3$

$x + y + z = 21$

Hence:

$P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P{X \le 1, Y = 17, Z = 3)$

This question implies that:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) + P(X = 1, Y = 17, Z = 3)$

Because

$0, 1 \le 1$ --- for x

In (e), we have:

$P(X = 1, Y = 17, Z = 3) = 0$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) +0$

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

In (a), we have:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

So:

$P(X=0; Y=17; Z = 3) = \frac{20!}{0! * 17! * 3!} * (5\%)^0 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{1 * 17! * 3!} * 1 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{17! * 3!} * (85\%)^{17} * (10\%)^{3}$

Expand

$P(X=0; Y=17; Z = 3) = \frac{20*19*18*17!}{17! * 3*2*1} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20*19*18}{6} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = 20*19*3 * (85\%)^{17} * (10\%)^{3}$

Using a calculator, we have:

$P(X=0; Y=17; Z = 3) = 0.07195$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

$P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1)$

This implies that:

$P(X \le 1) = P(X = 0) + P(X = 1)$

In (c), we established that X is a binomial distribution with the following parameters:

$p_1 = 5\%$ $n = 20$

Such that:

$P(X=x) = ^nC_x * p_1^x * (1 - p_1)^{n - x}$

So:

$P(X=0) = ^{20}C_0 * (5\%)^0 * (1 - 5\%)^{20 - 0}$

$P(X=0) = ^{20}C_0 * 1 * (1 - 5\%)^{20}$

$P(X=0) = 1 * 1 * (95\%)^{20}$

$P(X=0) = 0.3585$

$P(X=1) = ^{20}C_1 * (5\%)^1 * (1 - 5\%)^{20 - 1}$

$P(X=1) = 20 * (5\%)* (1 - 5\%)^{19}$

$P(X=1) = 0.3774$

So:

$P(X \le 1) = P(X = 0) + P(X = 1)$

$P(X \le 1) = 0.3585 + 0.3774$

$P(X \le 1) = 0.7359$

$(h)\ E(Y)$

Y has the following parameters

$p_2 = 85\%$ and $n = 20$

$E(Y) = p_2 * n$

$E(Y) = 85\% * 20$

$E(Y) = 17$

8 0

2 years ago

WILL MARK BRAINLIEST PLEASE HELP

Softa [21]

Answer:

$3 \: years \\ 30000 \times 0.97 \times 0.97 \times 0.97 = 27380.19 \\ 2 \: year \\ 30000 \times 0.97 \times 0.97 = 28227 \\ 1\: year \\ 30000 \times 0.97 = 29100$

5 0

3 years ago