<u>Complete Question</u>
The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. The results are: 59 for Bride, 50 for Groom, 30 for both. Find the probability that a randomly selected person from this sample was a friend of the bride OR of the groom.
Answer:
0.9875
Step-by-step explanation:
Total Number of Guests which forms the Sample Space, n(S)=80
Let the Event (a friend of the bride) =A
Let the Event (a friend of the groom) =B
n(A) =59
n(B)=50
Friends of both bride and groom, 
Therefore:
The number of Guests who was a friend of the bride OR of the groom = 79
Therefore:
The probability that a randomly selected person from this sample was a friend of the bride OR of the groom.
Answer:
Step-by-step explanation:
Y= 2.6x+49
1) 49
2)2.6
Answer:
A. 120 cubic meters
Step-by-step explanation:
Measurements in volume use <em>cubic</em><em> </em>units which deals with 3 dimensions
Answer:
So she can predict that she will be 39 out of 40 times she will be on time.
Step-by-step explanation:
The light rail service claims that it on time 98% of the time .
Given the girl takes light rail service 40 times per month .
Assuming the statement given by the light rail serve is correct 98% of 40 times the girl reaches on time , which is 39 times.
So she can assume that she will be 39 out of 40 times she will be on time.
No the light rails is not accurate if she was late 6 times out of 40 times in a month. Because by the statement given by light rail service she can miss a maximum of 1 time .
