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%
5,280 is greater because there is 1,760 yards in a mile.
Answer:
400 ft
Step-by-step explanation:
Let the distance traveled = d
Let the time taken = t
d α t^2
d = kt^2
k = d /t^2
When d = 100 and t = 2.5secs
k = 100/ (2.5)^2
k = 16
d = 16t^2
When t = 5secs
d = 16(5)^2
d = 16 * 25
d = 400 ft
27 inches ......... 3/8
x inches .............8/8
<h3>x = (27×8/8)/(3/8) = 27×8/3 = 216/3 = 72 inches (father)</h3>
I really sorry I have to do this I'm in a test I need help
