Let the width be W, then the length is 4W (since the width is 1/4 the length)
The area of the original deck is

The dimensions of the new deck are :
length = 4W+6
width=W+2
so the area of the new deck is :

"<span>the area of the new rectangular deck is 68 ft2 larger than the area of the original deck</span>" means that we write the equation:




the length is

ft
Answer: width: 4, length: 16
Answer:
Step-by-step explanation:
Given that a machine produces defective parts with three different probabilities depending on its state of repair.
condition Good order Wearing down Needs main Total
Prob 0.8 0.1 0.1 1
Defective 0.02 0.1 0.3
Joint prob 0.016 0.01 0.03 0.056
a) 0.016
b) total = 0.056
c) If not defective from needs maintenance
Prob for not defective = 
From machine that needs maintenance = 0.07
So reqd prob = 
Answer:Yes, he is correct
Step-by-step explanation:Plug 6 in as your x-value in the equation y=-275x+3500 and you will get y=1850
The expected length of code for one encoded symbol is

where
is the probability of picking the letter
, and
is the length of code needed to encode
.
is given to us, and we have

so that we expect a contribution of

bits to the code per encoded letter. For a string of length
, we would then expect
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have

so that the variance for the length such a string is

"squared" bits per encoded letter. For a string of length
, we would get
.