<u>Answer:</u>
X and Y are stochastically dependent RVs .
<u>Step-by-step explanation:</u>
Let ,
X = sum of the values that come up after throwing n (≥ 1) fare dice.
Y = number of times an odd number come up.
Let, n = 3
then, P(X =6) = p (say) clearly 0 < p < 1
and P (Y = 3) =
And,
P( X = 6, Y = 3) = 0 ≠
Hence, X and Y are stochastically dependent RVs
Answer:
Part A)
For 10 boxes
For 40 boxes
For 25 boxes
Part B) The number of boxes must be greater than or equal to 191 for a profit of at least $700
Part C) 151 boxes are needed to break even
Step-by-step explanation:
we know that
Profit, is equals to revenue minus costs of goods sold
we have
Part A) Find the profit for
1) 10 boxes
For x=10 boxes
substitute in the formula
The negative means that the revenue is less than the costs
2) 40 boxes
For x=40 boxes
substitute in the formula
The negative means that the revenue is less than the costs
3) 40 boxes
For x=25 boxes
substitute in the formula
The negative means that the revenue is less than the costs
Part B) How many boxes needed for a profit of at least $700?
For P=$700
substitute in the formula and solve for x
Round up
therefore
The number of boxes must be greater than or equal to 191 for a profit of at least $700
Part C) How many boxes needed to “break even”?
we know that
Break even is when the profit is equal to zero
For P=0
Round up
151 boxes are needed to break even
Hello there!
The question asks us to solve using the substitution method. The equations given are:
We are given a value for y, which is x-8. Substituting into the first equation of the system of equation gives us:
Use the distributive property to help simplify and solve for x
Combine like terms
Add 24 to both sides
Finally, divide both sides by 5
Now, let's use in the equation
to find the value of y
Now, we have the values x=5 and y=-3. Putting these values into an ordered pair gives us the answer (5,-3), which is the last answer choice.
Let me know if you need any clarifications, thanks!
~ Padoru