Which probability statements show that events A and B are independent? Select each correct statement.
OP(A) = 0.01, P(B) = 0.2, P(BA) = 0.2, and P(AB) = 0.01
O P(A) = 0.01, P(B) = 0.2, P(B|A) = 0.21, and P(A/B) = 0.21
OP(A) = 0.14, P(B) = 0.1, P(BA) = 0.1, and P(AB) = 0.14
O P(A) = 0.14, P(B) = 0.1, P(B|A) =W.24, and P(A|B) = 0.24

none of the option is correct.

step-by-step explanation:

it says to find which expression equals 9^3 sqrt 10 = ?

9^3 sqrt 10 = 729 sqrt 10.

checking option a. 5 sqrt 10 + 4 sqrt 10

5 sqrt 10 + 4 sqrt 10

= 9 sqrt 10.

checking option b. 5^3 sqrt 10 + 4^3 sqrt 10

5^3 sqrt 10 + 4^3 sqrt 10

= 125 sqrt 10 + 64 sqrt 10

= 189 sqrt 10.

checking option c. 5 sqrt 10 + 4^3 sqrt 10

5 sqrt 10 + 4^3 sqrt 10

= 5 sqrt 10 + 64 sqrt 10

= 69 sqrt 10.

checking option d. 5^3 sqrt 10 + 4 sqrt 10

5^3 sqrt 10 + 4 sqrt 10

= 125 sqrt 10 + 4 sqrt 10

= 129 sqrt 10.

hence, none of the option is correct.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

i think the answer is b.

step-by-step explanation:

not 100% sure though.