Factor the equation so...
(r^2-pr) and (p^2q-pqr)
Factor out (r^2-pr) = r(r-p)
Factor out (p^2q-pqr) = pq(p-r)
Add a negative to r(r-p) to make it -r(p-r)
(pq-r)(p-r) is the answer... I'm sorry I can't explain things well, but I tried.
Answer:
a_n = 2^(n - 1) 3^(3 - n)
Step-by-step explanation:
9,6,4,8/3,…
a1 = 3^2
a2 = 3 * 2
a3 = 2^2
As we can see, the 3 ^x is decreasing and the 2^ y is increasing
We need to play with the exponent in terms of n
Lets look at the exponent for the base of 2
a1 = 3^2 2^0
a2 = 3^1 2^1
a3 = 3^ 0 2^2
an = 3^ 2^(n-1)
I picked n-1 because that is where it starts 0
n = 1 (1-1) =0
n=2 (2-1) =1
n=3 (3-1) =2
Now we need to figure out the exponent for the 3 base
I will pick (3-n)
n =1 (3-1) =2
n =2 (3-2) =1
n=3 (3-3) =0
suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?
Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Mean = 177
standard deviation = 26
We find z-score using given mean and standard deviation
z = 
= 
=-0.11538
Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)
= 0.5438
P(weight will be greater than 174 lb) = 0.5438