When fritz drives to work his trip takes 3636 minutes, but when he takes the train it takes 2020 minutes. find the distance fritz travels to work if the train travels an average of 4848 miles per hour faster than his driving. assume that the train travels the same distance as the car?

To solve this problem, let us first assign the variables. Let us say that the trip when driving is denoted by 1, while when taking the train is denoted by 2.

So that we have variables of:

v1 or v2 = v = velocity of the trip

d1 or d2 = d = distance of the trip

t1 or t2 = t = time taken during the trip

The formula relating the three variables v, d and t is given as:

d = v t

Since the distance for each trip is equal therefore d1 = d2, so:

v1 t1 = v2 t2

It was also given that:

v2 = v1 + 48

Therefore:

v1 (36) = (v1 + 48) (20)

36 v1 = 20 v1 + 960

16 v1 = 960

v1 = 60 miles / hr

Therefore the distance is:

d1 = d = v1 t1

d = (60 miles / hr) (36 minutes) (1 hr / 60 minutes)

d = 36 miles

Therefore Fritz has to travel 36 miles going to work.

probability for j is 1/4

while

a is 1/6

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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