A football quarterback has two more chances to throw a touchdown before his team is forced to punt the ball. He misses the receiver on the first throw 25% of the time. When his first throw is incomplete, he misses the receiver on the second throw 15% of the time. What is the probability of not throwing the ball to a receiver on either throw? PLZ HELP!

A - the event he misses on the 1st throw,      P  ( A ) = 0.3

B - the event he misses on the 2nd throw.     P ( B/A ) = 0.1

If given 2 probabilities, the probability of both occurring is:

P ( A ) · P ( B/A ) = 0.3 · 0.1 = 0.03

The probability of not throwing the ball to a receiver on either throw is 0.03

(or 3 %).

3.75%

Step-by-step explanation:

Multiply the decimal points for the percentages and translate your answer into a percentage (percentages are out of 100, so you would move your answer 2 decimal places to the right) Hope this helps!

207/8 is the right

a = 3.14

b = 18

c = 18/5 or 3.6 or 3 3/5

step-by-step explanation: