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’m pretty sure it would be 4, since y= (x) which would be the numbers on the chart and then add 2
ex: y= (2) + 2
y= 4 and so on
Answer:
16
Step-by-step explanation:
10+6=16
Why?
The first thing we need to do is find the area of the triangle, we can to that by subtracting the area of ABCD from ACBE, then, we can use the formulas to calculate the area for both triangle and rectangle to find "f" and "g".
Calculating we have:
Now, we can calculate "f" by using the formula to calculate the area of the triangle:
Now, finding "g" by using the formula to calculate the area of the rectangle, we have:
Hence, we have that:
Have a nice day!
There are 60000 centigrams in 6 hectograms.
