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:
and?
Step-by-step explanation:
Answer:
- 6
Step-by-step explanation:
Given
y = 3(x - 1)(x + 2) ← expand factors using FOIL
= 3(x² + x - 2) ← distribute by 3
= 3x² + 3x - 6
To find the y- intercept let x = 0, thus
y = 3(0)² + 3(0) - 6 = 0 + 0 - 6 = - 6
Thus y- intercept = - 6 ⇒ (0, - 6 )
5.5% of 65 = 3.575
65 +3.575
= $68.575.
= $68.58.
The diameter is twice the radius (D=2R), then
