Question 19.). There is 1 + 1 + 3 + 4 + 3
Quiz Papers = 12
Four Papers with Scores of Six out Twelve
1 Selected:
Probability: 4/12
Answer: Letter Choice, (A), = 1/3
Question 20.). Two papers > 7
1 ===> 8, 1 ===> 9
Probability:
2/12
Answer: Letter Choice (F), 2/12 = 1/6
Hope that helps!!!! : )
Answer:
Most people found the probability of just stopping at the first light and the probability of just stopping at the second light and added them together. I'm just going to show another valid way to solve this problem. You can solve these kinds of problems whichever way you prefer.
There are three possibilities we need to consider:
Being stopped at both lights
Being stopped at neither light
Being stopped at exactly one light
The sum of the probabilities of all of the events has to be 1 because there is a 100% chance that one of these possibilities has to occur, so the probability of being stopped at exactly one light is 1 minus the probability of being stopped at both lights minus the probability of being stopped at neither.
Because the lights are independent, the probability of being stopped at both lights is just the probability of being stopped at the first light times the probability of being stopped at the second light. (0.4)(0.7) = 0.28
The probability of being stopped at neither is the probability of not being stopped at the first light, which is 1-0.4 or 0.6, times the probability of not being stopped at the second light, which is 1-0.7 or 0.3. (0.6)(0.3) = 0.18
Step-by-step explanation:
Answer:
= 4
Step-by-step explanation:
sqrt 15 = 3.872
to approximate to the nearest whole number (integer), you take the first decimal number which is 8, since 8 is greater than 5, you will add one(1) to the whole number which is 3 to make it 4.
Answer:
x=7 and m<LMN = 120
Step-by-step explanation:
if MO bisects LMN then 13x - 31 must be equal to x + 53
13x - x = 53 + 31
12x = 84
x = 7
and
13x - 31 + x + 53 = m<LMN
14x + 22 = m<LMN
since x is 7
14×7 + 22 = 120
Answer: Hello your question is incomplete attached below is the missing
n ( 1 + n )
Step-by-step explanation:
P( Bob hits target ) = 1/3
P( Eve hits target ) = 2/3
P( Carol hits target ) = 1
<u>Compute the P that Bob wins in a duel against Eve alone</u>
P(Bob hits the target in first shot ) = n = 1/3
P(Bob hits the target in second shot ) = n^2 = ( 1/3 * 1/3 ) = 1/9
hence the probability of Bob winning( i.e. P( Bob wins Event E1 ) = n + n^2 = n ( 1 + n )