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:
1458 sq cm
Step-by-step explanation:
because each dimension of A is 3 tomes each dimension of B, that will mean that both the length and width of B are multiplied by 3 in order to get the length and width of A, so the area of B is length×width=162 and the area of A (in terms of B's dimensions) is 3xlength×3×width=9×length×width=9×162=1458 sq cm
Answer:
pi will never end just trust me :p
Step-by-step explanation:
I’m pretty sure it’s d but i’m not 100%
A and C are correct, you have to multiply 32 by 4 and by 12, then add the aswers together to find how many juice boxes total.
