Answer:
0.18 ; 0.1875 ; No
Step-by-step explanation:
Let:
Person making the order = P
Other person = O
Gift wrapping = w
P(p) = 0.7 ; P(O) = 0.3 ; p(w|O) = 0.60 ; P(w|P) = 0.10
What is the probability that a randomly selected order will be a gift wrapped and sent to a person other than the person making the order?
Using the relation :
P(W|O) = P(WnO) / P(O)
P(WnO) = P(W|O) * P(O)
P(WnO) = 0.60 * 0.3 = 0.18
b. What is the probability that a randomly selected order will be gift wrapped?
P(W) = P(W|O) * P(O) + P(W|P) * P(P)
P(W) = (0.60 * 0.3) + (0.1 * 0.7)
P(W) = 0.18 + 0.07
P(W) = 0.1875
c. Is gift wrapping independent of the destination of the gifts? Justify your response statistically
No.
For independent events the occurrence of A does not impact the occurrence if the other.
The dimensions would be 40 m by 40 m.
To maximize area and minimize perimeter we make the dimensions as close to equal as possible. Since we only need fencing on 3 sides, 120/3 = 40; we can use 40 for each side.
B. Complementary means totaling 90 degrees, and supplementary means totaling 180 degrees.
Answer:
a. the number of people at Friday's, Saturday's, and Sunday's performances combined.
Step-by-step explanation:
The reason this information is needed to solve this problem is because we are looking to find the number of people who attended Saturday's performance. Without any information on Saturday's performance, there is no way to solve the problem. We are given the number of people on Friday and Sunday's performance, but are not given a total amount of people on all performances.
Answer:
after 75 minutes
Step-by-step explanation:
The least common multiple (LCM) of 15 and 25 is 75. It can be found a couple of ways:
1. List the factors of each number and find the product of the unique ones:
15 = 3·5
25 = 5²
The LCM is 3·5² = 75.
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2. Find the greatest common divisor (GCD) and divide the product of the numbers by that value. From the above list of factors, we see that 5 is the GCD of 15 and 25. Then the LCM is ...
15·25/5 = 75
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Or, you can simply list multiples of each number and see what the smallest number is that is in both lists:
15, 30, 45, 60, <em>75</em>, 90
25, 50, <em>75</em>, 100
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The two buses will appear together again after 75 minutes.
