The mean weight of gerbils is 1,200 g. The standard deviation is 50 g. Assuming a normal distribution, what percent of gerbils weigh above 1250 g? (ANSWER QUICK PLEASE)

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)

whatare you sayig

step-by-step explanation:

a) at most means it can cost $12 or less

so the inequality sign would be ≤

you would then need to multiply the number of toppings by the 0.35 for each one and add it to the cost of the cheese pizza.

the equation would be 9.99 + 0.35x ≤ 12

b) using the above equation:

9.99 + 0.35x ≤ 12

subtract 9.99 from each side:

0.35x ≤2.01

divide both sides by 0.35:

x ≤ 2.01 / 0.35

x ≤5.74

at most you could have 5 toppings.

the 5 topping pizza would cost: 0.35 x 5 = 1.75 + 9.99 = $11.74