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:
Approximate solution is 541.
Step-by-step explanation:
y' = 2xy, Δx = 0.4.
Make a table:
x y y' y' Δx + y
1 2 4 4*0.4 + 2 = 3.6 <---this is the new y value.
1.4 3.6 10.08 7.632
1.8 7.632 27.48 18.624
2.2 18.624 81.95 51.70
2.6 51.70 268.85 159.24
3 159.24 955.44 541.4
x-intercept = (15, 0) y-intercept = (0, 3)
slope (m) = (y₂ - y₁)/(x₂ - x₁) = (0 - 3)/(15 - 0) = -3/15 = -1/5
Point-Slope formula: y - y₁ = m(x - x₁) ; where (x₁, y₁) is one of the given points
y - 0 = (-1/5)(x - 15)
-5y = x - 15
x + 5y = 15
Answer: x + 5y = 15
Answer:
1.01 m8
Step-by-step explanation: