Answer: 0.0793
Step-by-step explanation:
Let the IQ of the educated adults be X then;
Assume X follows a normal distribution with mean 118 and standard deviation of 20.
This is a sampling question with sample size, n =200
To find the probability that the sample mean IQ is greater than 120:
P(X > 120) = 1 - P(X < 120)
Standardize the mean IQ using the sampling formula : Z = (X - μ) / σ/sqrt n
Where; X = sample mean IQ; μ =population mean IQ; σ = population standard deviation and n = sample size
Therefore, P(X>120) = 1 - P(Z < (120 - 118)/20/sqrt 200)
= 1 - P(Z< 1.41)
The P(Z<1.41) can then be obtained from the Z tables and the value is 0.9207
Thus; P(X< 120) = 1 - 0.9207
= 0.0793
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
a) because the denominators are the same add the numerators:
1 1/5 + 3 2/5 = 4 3/5
b) rewrite the fractions to have a common denominator:
1/2 = 3/6
1/3 = 2/6
Now subtract:
4 3/6 - 1 2/6 = 3 1/6
The answer choice A is correct
