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Mathematics, 21.05.2021 05:20 Kenastryker808

What is the coefficient of the x^4y^3 term in the expansion of (x-2y)^7? A. -280
B. -8
C. 560
D.35

Answers

ansver
Answer from: Demarcusclincy
The answer is -8jdjdh
ansver
Answer from: Quest

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

ansver
Answer from: Quest

answer: the correct option is d.

explanation:

it is given that the triangle j'k'l' shown on the grid below is a dilation of triangle jkl using the origin as the center of dilation.

from the given graph it is noticed that k(4,8) and k'(2,4). let origin is defined by o and scale factor is defined by k, then

k=\frac{ok'}{ok}     (1)

distance formula,

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

ok'=\sqrt{(2-0)^2+(4-0)^2}= \sqrt{4+16}= \sqrt{20}=2\sqrt{5}

ok=\sqrt{(4-0)^2+(8-0)^2}= \sqrt{16+64}= \sqrt{80}=4\sqrt{5}

put these values in equation (1).

k=\frac{2\sqrt{5}}{4\sqrt{5}}= \frac{1}{2}

therefore scale factor is   \frac{1}{2}   and option d is correct.

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What is the coefficient of the x^4y^3 term in the expansion of (x-2y)^7? A. -280
B. -8
...