A Ferris wheel is 40 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 34 meters above the ground?

t = 2.9517 min

Step-by-step explanation:

Given

D = 40 m ⇒ R = D/2 = 40 m/2 = 20 m

ybottom = 2 m

ytop =  ybottom + D = 40 m + 2 m = 42 m

yref = 34 m

t = 10 min

The height above the ground (y) is a sinusoidal function.

The minimum height is ybottom = 2 m

The maximum height is ytop = 42 m;

The midline is (ybottom + ytop)/2 = (2 m + 42 m)/2 = 22 m

If we model the wheel as follows

x² + y² = R²

where

y = yref - (R + ybottom) = 34 m - (20 m + 2 m) = 12 m

R = 20 m

we have

x² + (12 m)² = (20)²

⇒ x = 16 m

then

tan (θ/2) = x/y

⇒ tan (θ/2) = 16 m/12 m

⇒ θ = 106.26°

Knowing the angle of the circular sector, we apply the relation

t = (106.26°)*(10 min/360°)

⇒ t = 2.9517 min

Since the period of revolution is 10 minutes, the ride is above 34 meters for 2.9517 minutes each revolution.

the expected value of the random variable : 0, 1 and 2

step-by-step explanation: