Let x be the number of bags of Nature's Recipe Venison Meal & Rice Canine 20 percent protein, y be the number of bags of Nutro Max Natural Dog Food 27 percent protein and z be the number of bags of PetSmart Premier Oven Baked Lamb 25 percent protein a dog breeder buys. He wants to make 300 pounds of a mix. Then
20x+17.5y+30z=300.
Now
- in 20x pounds of Nature's Recipe Venison Meal & Rice Canine 20 percent protein is 0.2·20x=4x pounds of protein;
- in 17.5y pounds of Nutro Max Natural Dog Food 27 percent protein is 0.27·17.5y=4.725y pounds of protein;
- in 30z pounds of PetSmart Premier Oven Baked Lamb 25 percent protein is 0.25·30z=7.5z pounds of protein;
- in 300 pounds of mix containing 22 percent protein is 0.22·300=66 pounds of protein.
Then 4x+4.725y+7.5z=66.
You get a system

From the first equation
Substitute it into the second equation:

Simplify it:

y must be divided by 20, then
- y=0, x=9, z=10-6=4;
- y=20, x=16, z is not a whole number;
- y=40, x=23, z is not a whole number;
- y=60, x=30, z=10-20-35<0;
- thus, all other possible z will be <0.
Answer: 9 bags with 1st mix, 0 bags with 2nd mix and 4 bags with 3rd mix.
Answer:
a) 0.857
b) 0.571
c) 1
Step-by-step explanation:
Based on the data given, we have
- 18 juniors
- 10 seniors
- 6 female seniors
- 10-6 = 4 male seniors
- 12 junior males
- 18-12 = 6 junior female
- 6+6 = 12 female
- 4+12 = 16 male
- A total of 28 students
The probability of each union of events is obtained by summing the probabilities of the separated events and substracting the intersection. I will abbreviate female by F, junior by J, male by M, senior by S. We have
- P(J U F) = P(J) + P(F) - P(JF) = 18/28+12/28-6/28 = 24/28 = 0.857
- P(S U F) = P(S) + P(F) - P(SF) = 10/28 + 12/28 - 6/28 = 16/28 = 0.571
- P(J U S) = P(J) + P(S) - P(JS) = 18/28 + 10/28 - 0 = 1
Note that a student cant be Junior and Senior at the same time, so the probability of the combined event is 0. The probability of the union is 1 because every student is either Junior or Senior.