Answer:
66.48% of full-term babies are between 19 and 21 inches long at birth
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean length of 20.5 inches and a standard deviation of 0.90 inches.
This means that 
What percentage of full-term babies are between 19 and 21 inches long at birth?
The proportion is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19. Then
X = 21



has a p-value of 0.7123
X = 19



has a p-value of 0.0475
0.7123 - 0.0475 = 0.6648
0.6648*100% = 66.48%
66.48% of full-term babies are between 19 and 21 inches long at birth
To start with, we know there’s a definite amount of parking spaces in all(2,250) and we also know each level(6 in all) had 15 rows of parking spaces.
Equation
A = Parking spots per row
A = 2,250 Divided by(6 levels times 15 rows)
A= 2,250 Divided by 90
A = 25
The correct answer is 25 parking spots per row
To check this answer we can do
25 spots per row, Times 15 rows per level, Times the total amount of levels (6)
= the total amount of parking spots
25*15*6= 2,250
True
This confirms this answer as correct. Hope this helps.
36/27=1.3
7x1.3=9.1
The actual length is 9.1meters
You are looking for the average. To find the average, one adds all terms and divides by the number of terms.
((1/2)+(5/6))/2
((3/6)+(5/6))/2
(8/6)/2
(4/6)
(2/3)
The answer is B) 2/3.
1/3 of remained 2/3 is 2/9. We’ve used 1/3+2/9 = 3/9+2/9 = 5/9. So we’ve used 5/9 and we have 4/9 of the paint