Answer:
P(Y = 0) = 0.09
P(Y = 1) = 0.4
P(Y = 2) = 0.32
P(Y = 3) = 0.19
Step-by-step explanation:
Let the events be:
Wednesday
Thursday
Friday
Saturday
Their corresponding probabilities are

Since
= number of days beyond Wednesday that it takes for both magazines to arrive(so possible
values are 0, 1, 2 or 3)
The possible number of outcomes are therefore 
The values associated for each of the outcomes are as follows:

The probability mass function of
is,
![P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19](https://tex.z-dn.net/?f=P%28Y%20%3D%200%29%20%3D%200.3%280.3%29%20%3D%200.09%5C%5CP%28Y%20%3D%201%29%20%3D%20P%5B%28W%2C%20T%29%20or%20%28T%2C%20W%29%20or%20%28T%2C%20T%29%5D%5C%5C%3D%20%5B0.3%280.4%29%20%2B%200.3%280.4%29%20%2B%200.4%280.4%29%5D%5C%5C%3D%200.4%5C%5C%5C%5CP%28Y%20%3D%202%29%20%3D%20P%5B%28W%2C%20F%29%20or%20%28T%2C%20F%29%20or%20%28F%2C%20W%29%20or%20%28F%2C%20T%29%20or%20%28F%2C%20F%29%5D%5C%5C%3D%20%5B0.3%280.2%29%20%2B%200.4%280.2%29%20%2B%200.2%280.3%29%20%2B%200.2%280.4%29%20%2B%200.2%280.2%29%5D%5C%5C%3D%200.32%5C%5C%5C%5CP%28Y%20%3D%203%29%20%3D%20P%5B%28W%2C%20S%29%20or%20%28T%2C%20S%29%20or%20%28F%2C%20S%29%20or%20%28S%2C%20W%29%20or%20%28S%2C%20T%29%20or%20%28S%2C%20F%29%20or%20%28S%2C%20S%29%5D%5C%5C%3D%20%5B0.3%280.1%29%20%2B%200.4%280.1%29%20%2B%200.2%280.1%29%20%2B%200.1%280.3%29%200.1%280.4%29%20%2B%200.1%280.2%29%20%2B%200.1%280.1%29%5D%5C%5C%3D%200.19)
Step 1:
Calculate the measure of angle ∠ABC



From the triangle in the question,

Step 2:
Calculate the value of AB using the cosine rule below

By substituting the values, we will have
![\begin{gathered} b^2=a^2+c^2-2\times a\times c\times\cos B \\ b^2=10^2+15^2-2\times10\times15\times\cos 115^0 \\ b^2=100+225-300\times(-0.4226) \\ b^2=325+126.78 \\ b^2=451.78 \\ \text{Square root both sides} \\ \sqrt[]{b^2}=\sqrt[]{451.78} \\ b=21.26\operatorname{km} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20b%5E2%3Da%5E2%2Bc%5E2-2%5Ctimes%20a%5Ctimes%20c%5Ctimes%5Ccos%20B%20%5C%5C%20b%5E2%3D10%5E2%2B15%5E2-2%5Ctimes10%5Ctimes15%5Ctimes%5Ccos%20115%5E0%20%5C%5C%20b%5E2%3D100%2B225-300%5Ctimes%28-0.4226%29%20%5C%5C%20b%5E2%3D325%2B126.78%20%5C%5C%20b%5E2%3D451.78%20%5C%5C%20%5Ctext%7BSquare%20root%20both%20sides%7D%20%5C%5C%20%5Csqrt%5B%5D%7Bb%5E2%7D%3D%5Csqrt%5B%5D%7B451.78%7D%20%5C%5C%20b%3D21.26%5Coperatorname%7Bkm%7D%20%5Cend%7Bgathered%7D)
Hence,
The distance of point A to point C is = 21.26km
Here is for solving for x