Answer:
a on edge
Step-by-step explanation:
Answer:
Size
Step-by-step explanation:
Just got the answer out
Answer:
A= 0,2
B= 0,2
C= 0,4
D=0,2
Step-by-step explanation:
We know that only one team can win, so the sum of each probability of wining is one
P(A)+P(B)+P(C)+P(D)=1
then we Know that the probability of Team A and B are the same, so
P(A)=P(B)
And that the the probability that either team A or team C wins the tournament is 0.6, so P(A)+Pc)= 0,6, then P(C)= 0.6-P(A)
Also, we know that team C is twice as likely to win the tournament as team D, so P(C)= 2 P(D) so P(D) = P(C)/2= (0.6-P(A))/2
Now if we use the first formula:
P(A)+P(B)+P(C)+P(D)=1
P(A)+P(A)+0.6-P(A)+(0.6-P(A))/2=1
0,5 P(A)+0.9=1
0,5 P(A)= 0,1
P(A)= 0,2
P(B)= 0,2
P(C)=0,4
P(D)=0,2
22 is the answer because I just did it
Answer:
7.3% percentage of the bearings produced will not be acceptable.
Step-by-step explanation:
Consider the provided information.
Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.
Let X is the normal random variable which represents the diameter of bearing.
Thus, 0.500-0.004<X<0.500+0.004
0.496<X<0.504
The bearings have normally distributed diameters with mean value .499 inches and standard deviation .002 inches.
Use the Z score formula: 
Therefore



Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.
We need to find the percentage of the bearings produced will not be acceptable.
So subtract it from 1 as shown.
1-0.9270=0.073
Hence, 7.3% percentage of the bearings produced will not be acceptable.