Abaseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a speed of 26 ft/s.
(a) at what rate is his distance from second base decreasing when he is halfway to first base?
ft/s
(b) at what rate is his distance from third base increasing at the same moment?
ft/s

(a) Distance decreases at a rate of 11.628 ft/s.

(b) Distance increases at a rate of 11.628 ft/s.

Step-by-step explanation:

Refer to the diagram attached to at the bottom of this answer. We will call a the distance between the batter and 1st base, x the distance between the batter and 2nd base and y the distance between the batter and 3rd base.

The speed of the batter toward first base is 26 ft/s, which means:

Since the diamond is a square, the angle between trajectories in each base is a square angle, so we can use Pythagoras theorem:

Finding the first derivative:

When the batter is halfway to 1st, a = 45 ft, then:

So the distance between the batter and 2nd base decreases at 11.628 ft/s

Now, using Pythagoras again to find y:

and we know that

Finding the first derivative:

at  b = 45 ft

So the distance between the batter and 3rd base increases at 11.628 ft/s

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: