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, 10, 15
Step-by-step explanation:
the increase in the sequence is 5 so since it starts from 0 which you can find out because 25 is the 5th sequence and that means that +2*5 =25 . x=0
0+ 5 = sequence 1
0+ 5+5 = seq 2
0+5+5+5= seq 3
Answer:
a = 2
b = 3
c = 8
Step-by-step explanation:
First, let us look at the first two (far left) fractions to solve for a.
Since
, nothing needs to change in the numerator. 4 = 
a = 2.
Next, we will look at the first and third fractions to solve for b.
= 32
32 / 4 = 8
= 8
b = 3
Lastly, we will solve for c.
All of these fractions simplify to 8 and are equal to each other.
c = 8
Answer:
m = 2
n = 4
Step-by-step explanation:
Ok so to solve this, you want to get it so there is only one variable and then solve that equation. To do this, you can start by doing:
3m + n = 10
multiply both sides by 2
6m + 2n = 20
now, in both of your equations you have 2n so you can add the two equations:
6m + 2n = 20
+ 5m - 2n = 2
11m = 22
divide both sides by 11
m = 2.
Now, plug this value into the original equation, 3m + n = 10:
3 * 2 + n = 10
6 + n = 10
subtract 6 from both sides
n = 4