When shooting two consecutive free throws
during the regular season, a basketball player
makes the first free throw 78% of the time.
If he makes the first free throw, he makes
the second one 88% of the time, but he only
makes the second free throw 52% of the
time after missing the first one.
When he shoots a pair of free throws
in the team’s first playoff game, what
is the probability that he makes at least
one free throw?

89.44%.

Step-by-step explanation:

Let's work out the probability he misses both throws:

Prob( he misses both throws) =  (1-0.78) * ( 1 - 0.52)

= 0.22*0.48

= 0.1056.

So the probability he makes at least one free throw = 1 - 0.1056

= 0.8944.

(It is 1 - 0.1056 because the default of missing  both throws is either  making one throw  on first or second attempt, or making both throws).

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-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0