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:
Pentagon and I think the area is 19.89 so if you round it, that will be 19.9
Answer:
5/12
Step-by-step explanation:
You have to make it a common denomator and then subract until equeal.
Answer:
2.84×10²
Step-by-step explanation:
0.0284×10⁴
Move the decimal point 4 places to the right.
284
Now write in scientific notation by moving the decimal point 2 places to the left, behind the 2.
2.84×10²