Answer:
it's pretty hard right now for ne
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
Answer:
c. Mutation
Step-by-step explanation:
should be right I hope lol. good luck :)
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
