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%
9514 1404 393
Answer:
x = 4
Step-by-step explanation:
Divide by 2, subtract 8 ...
2(4/x +8) = 18
4/x +8 = 9
4/x = 1
4 = x . . . . . multiply by x
Answer:
1. b > -2
2. x <= 3
Step-by-step explanation:
Question 1:
-2(b + 5) < -6
Divide both sides by -2. Remember to change the inequality sign.
b + 5 > 3
Subtract 5 from both sides.
b > -2
Question 2:
-(x - 10) >= 7
Divide both sides by -1. Remember to change the inequality sign.
x - 10 <= -7
Add 10 to both sides.
x <= 3
7 oz bag:
110 ÷ 7 = 15.7143¢ per oz
9 oz bag:
146 ÷ 9 = 16.2222¢ per oz
So the answer is the 7 oz bag, hope this helps!
