# II100%A basketball coach collected data to analyze the free throw shooting percentages of players

###### Question:

A basketball coach collected data to analyze the free throw shooting percentages of players during a game and during practice. The

equation y = 0.9x - 1 is a line of best fit for the data, where x is the practice shooting percentage and y is the game shooting percentage.

Based on the equation, what will be the approximate shooting percentage in a game for a player with a practice shooting percentage of 88

?

OA. 78

B 80

C. 86

D. 99

## Answers

95% of the players loose.

The correct answer is option A; 78

Step-by-step explanation: The equation representing the line of best fit which is given as

y = 0.9x - 1

depicts the relationship between a player's practice results and actual game results. Whatever value is given as x which is the result from practice would be a good way of predicting a player's result when engaged in the real game.

For the question above, a player with a practice shooting percentage of 88 would have his game shooting estimated by simply inserting the value of his practice percentage into the equation showing the line of best fit shown as follows;

y = 0.9x - 1

Where x = 88

y = 0.9 (88) - 1

y = 79.2 - 1

y = 78.2

y ≈ 78

Therefore the approximate shooting percentage for a player with a practice shooting percentage of 88 would be 78 percent.

1.)49%

2.)5.3%

3)4.9%

4)46.1%

5)the events are independent of each other but not mutually exclusive.

6)4.9%

Step-by-step explanation:

From the data given

Event that a player in Npfl plays in his 30s is T

Event that a player plays in offense position is F

Drawing a set diagram

P(T) =10.2%

P(F)=48.6%

P(TnF)=4.9%

1.probability of T only or F only=10.2-4.9+48.6-4.9=49%

2.prob(Tonly)=prob(T)-Prob(PnT)

=10.2-4.9=5.3%

3.prob(TnF)=4.9%

4.Pr(TUF)'=100-{pr(Tonly)+Pr(Fonly)+P(TnF)}

100-(10.2-4.9+48.6-4.9+4.9)

100-(5.3+43.7+4.9)

=46 .1%

D.the two events are independent of each other i.e the occurrence of being in 30s and being an offensive players can happen simultaneously.

For thatsame reason ,they are NoT mutually exclusive.

E) prob (playing in 30s and being offensive) is p(TnF)=4.9%

20%

Step-by-step explanation:

One way to solve this is to subtract the number of boys from the total plyers to find the number of girls, then divide the number of girls by the total amount of players to find the percentage:

180 players - 108 boys = 72 girls.

72 girls / 180 players = 0.4 = 40% are girls.

Another way to solve is to find the percent of boys and subtract that from 100%:

108 boys / 180 player = 0.6 = 60% are boys

100% - 60% = 40% are girls.

The answer is 19%

Step-by-step explanation:

Due to the lack of the box plot, this is hard to answer. I do want to say that I hope none of the players weight greater than or equal to 194194 pounds.

If you want a realistic answer, it is 100%. Other than that, there is not enough information.

It is 95% of people that don't win

The first thing you are gonna do is, subtract the number of students (180) to the number of boys (108). 180-108= 72. So you got 72. Then divide the 72 to number of students (180). 72/180= 0.4. After that multiply the 0.4 by 100 0.4 x 100= 40%.

40% of the players that signed up were girls