if im not mistaken it is 6.05 because if 3/4 (0.75) is the weight of pears, you subtract that by the total weight and the remanding is your weightof apples
Answer:
1: 0.027
2: 0.01375
Step-by-step explanation:
Answer:
Expected value of profit is ≅ $15.80
Step-by-step explanation:
from the Question,
Let the variable X represents the expected value of profit. X is called as random variable because picking a number from 000-999 digits is a Random process.
P(win) = ![0.001](https://tex.z-dn.net/?f=0.001)
So, P(lose) = ![1-0.001=0.999](https://tex.z-dn.net/?f=1-0.001%3D0.999)
Suppose that,
we really want to win this lottery. so we can go to the store and spend $1000 to buy all ticket (from 000 - 999). This would ensure your winning of $500 with one of the tickets (for a $499 profit), but the other 999 would be losers (for a $999 loss).
What would be your average winnings on a per-ticket basis?
∑![x\times P(x)](https://tex.z-dn.net/?f=x%5Ctimes%20P%28x%29)
![= 499\times0.0001+ (-1)\times0.999](https://tex.z-dn.net/?f=%3D%20499%5Ctimes0.0001%2B%20%28-1%29%5Ctimes0.999)
= ![-0.50](https://tex.z-dn.net/?f=-0.50)
Here,
Standard deviation of the expected winnings
∑![(x-u)^{2} \times P(x)](https://tex.z-dn.net/?f=%28x-u%29%5E%7B2%7D%20%5Ctimes%20P%28x%29)
= ∑ ![(499-(-0.50))^{2} \times 0.001 + (-1-(-0.50))^{2} \times 0.999](https://tex.z-dn.net/?f=%28499-%28-0.50%29%29%5E%7B2%7D%20%5Ctimes%200.001%20%2B%20%28-1-%28-0.50%29%29%5E%7B2%7D%20%5Ctimes%200.999)
![= 249.75](https://tex.z-dn.net/?f=%3D%20249.75)
Taking square root of the variance to get the standard deviation:
SD(x) = ![\sqrt{249.75}](https://tex.z-dn.net/?f=%5Csqrt%7B249.75%7D)
≅ $15.80
Hence
The expected value of profit is ≅ $15.80
Answer:
x = 10
Step-by-step explanation:
Given the 2 equations
x - y = 11 → (1)
2x + y = 19 → (2)
Adding (1) and (2) term by term will eliminate the y- term
(2x + x) + (y - y) = (19 + 11)
3x = 30 ( divide both sides by 3 )
x = 10
$328.99 x .20 = $65.79
$328.99 - $65.79 = $263.20