10. A 11. D 12. A (wasn’t too sure about question 12)
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%
It’s 0.09 or .09. Hope this helps:)
Answer:
91
Step-by-step explanation:
Look at multiples of 7 and you’ll find 91.
91 dived by 3 is 30 with a remainder of 1.
91 divided by 5 is 18 with a remainder of 1.
91 divid by 7 is 13 with no remainder.
