Answer:
probability that both passes a defective item is 0.8742
<h3>
Step-by-step explanation:</h3>
probability that the first inspector misses is Pr( 1st misses)= 0.06
therefore the probability he does not miss is
Pr(1st passes)= 1 - Pr( 1st misses) = 1 - 0.06 = 0.94
probability that the second misses is Pr( 2nd misses) = 0.07
therefore probability that 2nd does not miss is
Pr( 2nd passes) = 1- Pr( 2nd misses) = 0.93
probability that both passes a defective item is Pr(1st passes)*Pr( 2nd passes)
= 0.93*0.94 = 0.8742
To answer this question, we need to find the winning probability in either case.
Probability = no. of outcomes / total no. of possible outcomes
<u>When Hope pulled her defender :</u>
Total no. of games = 9
No. of games won = 3
Winning probability = 3/9 =1/3
<u>When Hope left her defender :</u>
Total no. of games = 10
No. of games won = 6
Winning probability = 6/10 = 3/5
We know that , 1/3 < 3/5.
So, Hope should not pull her defender, as the winning probability is better when Hope left her defender.
Answer : A. Hope should not pull her defender.
1q+2=-6q+3+8q+4
1q+2=2q+7
<span>-2q -2q
</span>-q+2=7
<span> -2 -2
</span>-q=<span>5
-1 -1
q=-5
</span>
Answer:
28 and 15
Step-by-step explanation:
Answer:
n≅376
So sample size is 376.
Step-by-step explanation:
The formula we are going to use is:
where:
n is the sample size
p is the probability of favor
q is the probability of not in favor
E is the Margin of error
z is the distribution
α=1-0.98=0.02
α/2=0.01
From cumulative standard Normal Distribution
p is taken 0.5 for least biased estimate, q=1-p=0.5
n≅376
So sample size is 376
