25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Answer:
27
Step-by-step explanation:
r = 9
s = 6
r^2
= r x r
= 9 x 9
= 81
13s
= 13 x 6
= 108
13s - r^2
= 108 - 81
= 27
2.5 what just use the formula pi•(4/3r•r)
It is simple, you are converting 150 of 200 into a fraction and reducing it
150/200=15/20
5 goes into 20 4 times
15/20=1/4
1/4=0.75=75%
A) the only mistake is changing the mixed number to an improper fraction .
