We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.
The probability of a data score between two z-scores is given by formula 
.
Using above formula, we will get:
Now we will use normal distribution table to find probability corresponding to both z-scores as:
Now we will convert 
into percentage as:
Upon rounding to nearest tenth of percent, we will get:
Therefore, our required probability is 27.4% and option C is the correct choice.
(-2)^3
12^0
I don't know the rest
I added a screenshot with the complete question
Answer:x = 3
y = 9
Explanation:1- getting the value of x:We are given that:
side AB is congruent to side DF. This means that:
AB = DF
3(2x+10) = 12x + 12
6x + 30 = 12x + 12
12x - 6x = 30 - 12
6x = 18
x = 18/6
x = 3
2- getting the value of y:We are given that:
side BC is congruent to side FG. This means that:
BC = FG
2y + 12 = 2(2y-3)
2y + 12 = 4y - 6
4y - 2y = 12 + 6
2y = 18
y = 18/2
y = 9
Hope this helps :)
Answer: x = 6V/bh
Explanation:
V = 1/6 bhx
V = (bh/6)x
V • 6/bh = x
6V/bh = x
