2t^2 =4t+3t=5t
5t+500
3t^2=9t+t=10t
10t+300
Answer:
0.018 is the probability that a randomly selected college student has an IQ greater than 131.5
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 15
We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.
Formula:
a) P(IQ greater than 131.5)
P(x > 131.5)
Calculation the value from standard normal z table, we have,
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
The missing side length is 7 yds.
