I’m not sure but I think 8
Answer: (8y+1)(y+5)
Step by step:
8y2 + y +40y + 5
y(8y+1) + 5(8y+1)
(8y+1)(y+5)
a) For x = 27:
z = 27-28/2 = -0.5
For x = 31:
z = 31-38/2 = 1.5
From the normal distribution table, P(27 < x < 31) = P(-0.5 < z < 1.5) = P(z < 1.5) - P(z < -0.5) = 0.9332 - 0.3085 = 62.47%
b) For x > 30.2:
z = 30.2-28/2 = 1.1
From the normal distribution table, P(x > 30.2) = P(z > 1.1) = 1 - P(z > 1.1) = 1 - 0.8643 = 13.57%
Answer:
3
Step-by-step explanation:
The boxes are stacked 5 boxes deep by 4 boxes high by 4 boxes across, then there are
boxes in total.
The mass of 1 box of paper is 22.5 kilograms, so 80 boxes weigh
kilograms.
When the driver is in the truck, the mass is 2948.35 kilograms, then the total mass is

Let n be the number of boxes of paper the driver must deliver at the first stop. Their weigth is 22.5n kg and the weight of the truck without n boxes is

Trucks with a mass greater than 4700 kilograms are not allowed over the bridge, thus

Hence, the driver must deliver at least 3 boxes at the first shop.
Factorize the numerator and denominator. You'll see that they both have a factor of 4 that can be canceled. The introduce a factor of 3 to change the denominator to 36.
