Answer:
Throw a die
Every number on the die represents one of the 6 prizes
The amount of times you throw the die describes the number of boxes you need.
Step-by-step explanation:
We'll represent the distance as d and the time as t.
d= 3(t+7) because Gilda is 7 minutes late when travelling 3 mph
d= 4(t-5) because Gilda is 5 minutes early at 4 mph
d= 3t + 21
d= 4t -20
3t +21 =4t -20
41=t
d= 3(41) +21= 144
The distance is 144 miles.
A coin has two sides, which means it has a 1/2 chance of landing on each individual side when you flip the coin. Since there is a 1/2 chance of landing on each side, you could expect the coin to land on "Heads' 75 times.
The definition of the set E gives you a natural choice for the limits in the integral:

Computing the integral, we get


