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mrs_skeptik [129]
3 years ago
6

Match each equation to equivalent equation. -4x+3y=20 -6x-2y=10. 2x +5y=18

Mathematics
1 answer:
alex41 [277]3 years ago
6 0

Answer:

4x + 10y = 36  ≡  to 2x + 5y = 18

As you can see, the the coefficients on the first one are double those in the second.

12x - 9y = -60  ≡  to -4x + 3y = 20

In this case, the terms in the first one are all the same, but multiplied by -3

And by simple elimination we can say:

-24x - 8y = 40 ≡ -6x - 2y = 10

In this case, the terms of the first equation are equal to those of the latter, but all multiplied by 4.

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Verify that the functions are probability mass functions, and determine the requested probabilities.
Free_Kalibri [48]

Answer:

a) 3/64 = 0.046 (4.6%)

b) 63/64 = 0.9843 (98.43%)

c) 1/64 = 0.015 (1.5%)

d) 1/4 = 0.25 (25%)

Step-by-step explanation:

in order to verify that the f(x) is a probability mass function , then it should comply the requirement that the sum of probabilities over the entire space of x is equal to 1. Then

∑f(x)*Δx = 1

if f(x)=(3/4)(1/4)^x , x = 0, 1, 2, ...

then Δx=1 and

∑f(x) = (3/4)∑(1/4)^x = (3/4)* [ 1/(1-1/4)] = (3/4)*(4/3) = 1

then f represents a probability mass function

a) P(X = 2)= f(x=2) = (3/4)(1/4)^2 = 3/64 = 0.046 (4.6%)

b) P(X ≤ 2) = ∑f(x) =  f(x=0)+ f(x=1) + f(x=2) = (3/4) + (3/4)(1/4) +  3/64 = 63/64 = 0.9843 (98.43%)

c) P(X > 2)= 1- P(X ≤ 2) = 1 - 63/64 = 1/64 = 0.015 (1.5%)

d) P(X ≥ 1) = 1 - P(X < 1) = 1 - f(x=0) = 1- 3/4 = 1/4 = 0.25 (25%)

4 0
3 years ago
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Answer:

A

Step-by-step explanation:

a and d look similar but I'm quite sure it's a

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Answer:

its D

Step-by-step explanation:

3 0
2 years ago
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For example 9.............
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3 years ago
Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(2x)^5(+3)^0\\&#10;2&+5&(2x)^4(+3)^1\\&#10;3&+10&(2x)^3(+3)^2\\&#10;4&+10&(2x)^2(+3)^3\\&#10;5&+5&(2x)^1(+3)^4\\&#10;6&+1&(2x)^0(+3)^5&#10;\end{array}

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for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(2x)^4(-3y)^0\\&#10;2&+4&(2x)^3(-3y)^1\\&#10;3&+6&(2x)^2(-3y)^2\\&#10;4&+4&(2x)^1(-3y)^3\\&#10;5&+1&(2x)^0(-3y)^4&#10;\end{array}


\bf (3a+4b)^8\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(3a)^8(+4b)^0\\&#10;2&+8&(3a)^7(+4b)^1\\&#10;3&+28&(3a)^6(+4b)^2\\&#10;4&+56&(3a)^5(+4b)^3\\&#10;5&+70&(3a)^4(+4b)^4\\&#10;6&+56&(3a)^3(+4b)^5\\&#10;7&+28&(3a)^2(+4b)^6\\&#10;8&+8&(3a)^1(+4b)^7\\&#10;9&+1&(3a)^0(+4b)^8&#10;\end{array}

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
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3 years ago
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