The area of a rectangle is 77 ft2 , and the length of the rectangle is 10 ft less than three times the width. find the dimensions of the rectangle.

width (w) = w

Length (L) = 3w - 10

Area (A) = L x w

77 = (3w - 10)w

77 = 3w² - 10w

0 = 3w² - 10w - 77

0 = 3w² - 21w + 11w - 77

0 = 3w(w - 7) + 11(w - 7)

0 = (3w + 11)(w - 7)

0 = 3w + 11, 0 = w - 7

w = , w = 7 (width cannot be negative!)

w = 7

Length (L) = 3w - 10 = 3(7) - 10 = 21 - 10 = 11

width = 7 ft, Length = 11 ft

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! )

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.