Resolving Probabilities of Exactly One and At Least One: Suppose the US Transportation Security Administration (TSA) requires 10% of all airplane passengers be randomly selected for a full screening before boarding a plane. Now, suppose you will travel on a plane 8 times in the coming year. (a) What is the probability that you will get screened exactly once?
(b) What is the probability that you will get screened at least once?
(c) Is the probability in favor (> 50%) of you getting screened exactly once? Is the probability in favor (> 50%) of you getting screened at least once?

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!

for this case we must apply distributive property to the following expression:

the distributive property states that:

then, when applying the property we have to multiply the -2 for each term within the parenthesis:

step-by-step explanation:

hello 0xxbadwolfxx0,

2.6 x 10^9 is 2,600,000,000.

1.3 x 10^2 is 130

multiplying 130 and 2,600,000,000 would simply give you 200,000,000,000.

therefore, the answer would be d. 2.0 × 10^11.

i hope i ! warm regards, mae.