Step-by-step explanation:
6. 7. 13
18. 10. 28
24. 17. 41
4/6 with 2 throws left i think.
Answer:
0.03 is the probability that for the sample mean IQ score is greater than 103.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 16
Sample size, n = 100
We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =

P( mean IQ score is greater than 103)
P(x > 103)
Calculation the value from standard normal z table, we have,

0.03 is the probability that for the sample mean IQ score is greater than 103.
Answer:
Step-by-step explanation:
2a + 6 / a² -9 = 2(a + 3) / (a² -3²)
= 2(a+3) / (a+3)(a-3) {a²-b² = (a+b)(a-b)
= 2 / (a-3)