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 two at the top technically means how many times the number in the bracket times ITSELF by, so its technically -3 x -3 which is +9 , cause a negative x negative = positive
Answer:
LCD(8/
15
, 11/
30
, 3/
5
) = LCM(15, 30, 5) = 2×3×5 = 30
Step-by-step explanation:
5,740,000( you have to make the number less than 10
5.740000 ( you put a decimal to make it less )
5.74e6 (you count how many times you skipped over to get at the decimal point , and that's your awnser .
