All categories

4 years ago

WILL GIVE BRAINLIEST :)

1 answer:

7 0

Step One
Develop Pascal's Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1 <<<< How the 5th power expands.

Step Two
Confirm that nCr gives the same thing as the triangle gives.

The third term is outlined as 10 a^3 b^2 The powers must add to 5.
The 10 can be found this way

nCr Where n is the row number and r is the term you want so
n = 5
r = 3
nCr = n! / [(n - r)! r!]
5C3 = 5! / ((3!) (2!) )
5C3 = 5 * 4 / 2
5C3 = 10 Just as the triangle predicted.

Step Three
Find a^3 b^2
a = 3x
b = - 5y

a^3 = (3x)^3
a^3 = 27x^3
b^2 = (-5y)^2
b^2 = 25 y^2

Step Three
Find the Answer
Third term = 10*27x^3*25y^2
Third term = 6750 x^3y^2

You might be interested in

Suppose in the population, the Anger-Out score for men is two points higher than it is for women. The population variances for m

tatuchka [14]

Answer:

a)= 2

b) 6.324

c) P= 0.1217

Step-by-step explanation:

a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)

b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by

σ_X`1- X`2 = √σ²/n1 +σ²/n2

Var ( X`1- X`2) = Var X`1 + Var X`2 = σ²/n1 +σ²/n2

σ_X`1- X`2 =√20 +20 = 6.324

if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.

Where as Z is normally distributed with mean zero and unit variance.

If we take X`1- X`2= 0 and u1-u2= 2 and standard deviation of the sampling distribution = 6.324 then

Z= 0-2/ 6.342= -0.31625

P(-0.31625<z<0)= 0.1217

The probability would be 0.1217

8 0

3 years ago

PLEASE HELP QWQ AsAp with these 4 questions

den301095 [7]

Answer:

Step-by-step explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

$\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]$ and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

$(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)]$ which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

$\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right]$ and multiplying gives us

$(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)]$ which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

$\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right]$ and multiplying gives us

$(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)]$ which gives us:

$-126+48-90-[-280+54-36]$ which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

$\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right]$ $\left[\begin{array}{c}x\\y\\\end{array}\right]$ = $\left[\begin{array}{c}-5\\-2\\\end{array}\right]$ and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

$\frac{1}{2}\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]$ (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

$\left[\begin{array}{cc}-4&\frac{5}{2}\\3&-2\\\end{array}\right]$ Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

$\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&\frac{5}{2} \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right]$ and that right side multiplies out to