I NEED HELP ASAP A football quarterback goes for a two-point conversion when the ball is within 10 yards of the end zone. During the game, he has two opportunities for a two-point conversion. He misses the first attempt 50% of the time. When he misses the first attempt, he misses the second attempt 15% of the time. What is the probability of missing both two-point conversion attempts?

P = 0.075     P = 7.5%

Step-by-step explanation:

We pretty much look for the probability that the player fails the first attempt and also fails the second attempt.

We know for a fact that the probability that the quarterback fails the first attempt is:

p(1) = 0.5

and we know that the probability that the quarterback fails the second attempt after failing the first is:

p(2/1) = 0.15  

Then the probability of failure of both attempts is:

P(2/1) = p(1 and 2) / p(1)

p(1 and 2) = P(1) * P(1/2)

p(1 and 2) = 0.5 * 0.15

and that is

p(1 and 2) = 0.075

you probably didn't want the nerdy stuff but oh well

x+14

step-by-step explanation:

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.