1/5 five being the denominator, because in math you must always simplify
Answer:
We accept H₀
Step-by-step explanation:
Normal Distribution
size sample n = 69
sample mean 18.94
standard deviation 8.3
Is a one tailed-test to the left we are traying of find out is we have enough evidence to say that the mean is less than 20 min.
1.-Test hypothesis H₀ ⇒ μ₀ = 20
Alternative hypothesis Hₐ ⇒ μ₀ < 20
2.- Critical value
for α = 0.1 we find from z Table
z(c) = - 1.28
3.-We compute z(s)
z(s) = [ ( μ - μ₀ ) / (σ/√n) ⇒ z(s) = [( 18.94 - 20 )*√69)/8.3]
z(s) = ( -1.06)*8.31/8.3
z(s) = - 1.061
4.- We compare
z(c) and z(s) -1.28 > -1.061
Then z(c) > z(s)
z(s) in inside acceptance region so we accept H₀
Answer:
B.
Step-by-step explanation:
Well we know that
so we can get the 2 outside of the radical
and we can get the x^2 outside too.
and we also can get y outside.
so we have:
![2x^{2}y\sqrt[5]{7xy^3}](https://tex.z-dn.net/?f=2x%5E%7B2%7Dy%5Csqrt%5B5%5D%7B7xy%5E3%7D)
Answer:
Step-by-step explanation:
A.f(x) =(x-1)^2 is even because of that even exponent, 2.
b. F(x)=8x is odd because x has the exponent 1.
c.f(x) =x^2-x is neither even nor odd
d.f(x)=7 is even because the exponent is even: 7^0
