Civil engineers believe that w, the amount of weight (in units of 1000 pounds) that a certain span of a bridge can withstand without structural dam- age resulting, is normally distributed with mean 400 and standard deviation 40. suppose that the weight (again, in units of 1000 pounds) of a car is a random variable with mean 3 and standard deviation .3. approximately how many cars would have to be on the bridge span for the probability of structural damage to exceed .1?

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation:

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

answer: $201

step-by-step explanation: to get the answer, you first get 3 nieces that he sold and multiply it by 67 and get $201, so he earned $201 selling his nieces.