Vugar has a maximum of 30 batteries for toys. each toy helicopter uses the same number of batteries, and each toy car uses the same number of batteries. let h represent the number of toy helicopters and c represent the number of toy cars that vugar can play with without running out of batteries. 3h+4c? 303 according to the inequality, how many batteries does each toy helicopter use, and how many batteries does each toy car use? does vugar have enough batteries to play with 5 toy helicopters and 4 toy cars? choose 1 a) yes b) no

Each toy helicopter uses 3 batteries, and each toy car uses 4 batteries.

B) No, Vugar does not have enough batteries to play with 5 toy helicopters and 4 toy cars

Probability of a student playing both basketball and baseball is 3/28

Step-by-step explanation:

Step 1:

It is given the class has 28 students out of which 17 play basketball and 9 play baseball. It is also given that 5 students play neither sport.

Total number of students = 28

Students playing neither sport = 5

Students playing atleast one sport = 28 - 5 = 23

Step 2:

Let N(Basketball) denote the number of students playing  basketball and N(Baseball) denote the number of people playing  baseball.

Then N(Basketball U Baseball) denotes the total number of students playing basketball and baseball and N(Basketball ∩ Baseball) denotes playing both basketball and baseball.

Since the number of students playing atleast one sport is 23, N (Basketball U Baseball) = 23.

N (Basketball U Baseball) = N(Basketball) + N(Baseball) - N(Basketball ∩ Baseball)

N(Basketball ∩ Baseball) = N(Basketball) + N(Baseball) - N (Basketball U Baseball)

N(Basketball ∩ Baseball) = 17 + 9 - 23 = 3

Step 3:

Number of students playing both basketball and baseball = 3

Total number of students = 28

Probability of a student playing both basketball and baseball is 3/28

Step 4:

Probability of a student playing both basketball and baseball is 3/28