Answer:
this question is very unorganized please make it more clear, thanks :)
Step-by-step explanation:
6 = 18
—. ——
3. 9
This is for a
__
Here,
let width(b)be x then,
length (l)=9cm+x
area =112 sq cm
now,
area of rectangle=l*b
or, 112=(9+x)x
or, 112=9x+x^2
or, 0=x^2+9x-112
or, 0=x^2+(16-7)x-112
or, 0=x^2+16x-7x-112
or, 0=x(x+16)-7(x+16)
or, 0=(x-7)(x+16)
either,
0=x-7
or,7=x
x=7cm
Or,
0=x+16
or, -16=x
x= -16[impossible,as distance is never negative] so,
x=7cm
therefore,length = 7cm + 9 cm = 16cm and width = 7cm.
;)
Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if 
and 
then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.
side note: q = 1-p is the complement of probability p
The first one is not proportional and the second one is proportional
