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.
I would not normally stop to answer this question, mainly because
there is no question asked. It's just three statements.
I do have to stop here and leave a remark, however.
Math and Physics are closely enough related that I would not
have expected to see what I see here. Although the math in
this question is reasonable, the Physics is inexcusable.
A rainbow is always a part of a circle.
Rainbows are never parabolas.
You could never cut a parabola out of paper, and then
hold it up in front of you after a rainstorm, and match it
to the rainbow.
Answer:
See below
Step-by-step explanation:
Answer:
4sqrt(2) cm
Step-by-step explanation:
a^2 + b^2 = c^2
4^2 + 4^2 = c^2
32 = c^2
sqrt(32) = c
c = 4sqrt(2)
8:15 is the answer assuming the clock is at PM.
