D. When you do the math, the results to everything should be t=9. D is the only outlier. Therefore, D is the answer in this case.
Answer:
88.51 is the minimum score needed to receive a grade of A.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 73
Standard Deviation, σ = 11
We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.0793.
Calculation the value from standard normal z table, we have,
Hence, 88.51 is the minimum score needed to receive a grade of A.
Answer:
-5/2+-1/2√37≤x≤-5/2+1/2√37
Step-by-step explanation:
Step 1: Find the critical points
-x^2-5x+3=0
For this equation: a=-1, b=-5, c=3
−1x^2+−5x+3=0
x=−b±√b2−4ac/2a
x=−(−5)±√(−5)2−4(−1)(3)/2(-1)
x=5±√37
/−2
x=-5/2+1/2√37
Step 2: Check intervals in between critical points
x≤-5/2+1/2 √37 (Doesn't work in original inequality)
-5/2+-1/2√37≤x≤-5/2+1/2√37 (Works in original inequality)
x≥-5/2+1/2 √37 (Doesn't work in original inequality)
Answer:
a. 254.34, b. 56.52
Step-by-step explanation:
a. A = πr^2,
A = 3.14(9^2)
A = 3.14*81
A = 254.34
b. C = πd
C = 18*3.14
C = 56.52
The volume of the pyramid is 15 yd
