Answer:
A.0.4477
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that a randomly selected exam will require between 14 and 19 minutes to grade?
This probability is the pvalue of Z when X = 19 subtracted by the pvalue of Z when X = 14. So
X = 19
has a pvalue of 0.7389.
X = 14
has a pvalue of 0.2912
0.7389 - 0.2912 = 0.4477
So the correct answer is:
A.0.4477
False - Correct on apex!!!
Part 1.
A is.
45 and 34 wood pieces left
B is.
It takes 11 cause 100-89=11 and 11+89=100 and it keeps subtracting 11.
C is.
I dont understand the question(sorry)
D is.
At least 9 shelves,
100-11=89
89-11=78
78-11=67
67-11=56
56-11=45
45-11=34
34-11=23
23-11=12
12-11=1
And you would have 1 wood piece remaining!!
Answer:
Amount of hay in first barn = 90 tons
Amount of hay in second barn song = 30 tons
Step-by-step explanation:
Let the amount of hay in second barn be x, then hay in first barn will be 3x.
20 tons of hay was removed from first barn and ws added to second barn.
Now, the amount of hay in first barn = 3x -20
Amount of hay in second barn = x+20
The equation now formed is:
x+20 = 
(3x-20)
7x+14 = 15x -100
-8x = -240
8x = 240
x = 30
3x = 90
Amount of hay in first barn = 90 tons
Amount of hay in second barn song = 30 tons
