At a local restaurant, 40% of the lunch customers order a hamburger, 70% of the lunch customers order French fries, and 30% of lunch customers order both a hamburger and French fries. If a lunch customer is randomly selected from this restaurant, what is the probability that the customer order fries if we know they ordered a hamburger? 0.75

0.43

0.40

0.30

the fist one is not equalvalent but i think th second one isand the third isnt definetly or the last one is i think. hope this !

step-by-step explanation:

9 sandwiches per hour

step-by-step explanation:

when we have a rate, we want a 1 in the denominator

27/3 = 9

27 sandwiches/ 3 hours = 9 sandwiches/hour

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: