This for some reason is in the middle school I'll move that for you and y = 17 I think
Answer:
The solution is:
z (max ) = 446.25 $
x₁ = 55 x₂ = 0 x₃ = 5 x₄ = 70
Step-by-step explanation:
From problem statement:
Nacholupa Quesatilla Enchinacho Burritaco
x₁ x₂ x₃ x₄
Cheese 2 4 2 4
meat 0 2 2 2
beans 4 0 2 0
Tortillas 3 1 3 1
Price $ 2.75 2 3 4
From table :
z = 2.75*x₁ + 2*x₂ + 3*x₃ + 4*x₄ to maximize
Subject to:
Availability of cheese: 400 ou
2*x₁ + 4*x₂ + 2*x₃ + 4*x₄ ≤ 400
Availability of meat : 150 ou
2*x₂ + 2*x₃ + 2*x₄ ≤ 150
Availability of beans : 400 ou
4*x₁ + 2*x₃ ≤ 400
Availability of tortillas : 250
3*x₁ + x₂ + 3*x₃ + x₄ ≤ 250
General constraints:
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 x₄ ≥ 0
All integers
With the use of AtomZmath ( integer solving on-line software )
The solution is:
z (max ) = 446.25 $
x₁ = 55 x₂ = 0 x₃ = 5 x₄ = 70
Answer:
If you buy the paint set and the 6 canvases its B if you just buy the cancases itd D
Step-by-step explanation:
Using a calculator with the binompdf and binomcdf features, I can calculate these values. My calculator is a TI-83 plus, and the features are found under the 2nd, Vars keys (Scroll up or down until you see them).
If "exactly" is to be found, use binompdf:
binompdf(number of trials, probability of success, exactly number)
ANSWER for exactly 3: binompdf(8, 0.5, 3) = 0.21875 = 21.875%
If "at least" is to be found, use binomcdf:
binomcdf(number of trials, probability of success, at least number - 1)
ANSWER for at least 6: binomcdf(8, 1/2, 5) ≈ 0.8555 ≈ 85.55%
If "at most" is to be found, use binomcdf:
binomcdf(number of trials, probability of success, at most number)
ANSWER for at most 3: binomcdf(8, 0.5, 3) ≈ 0.3633 ≈ 36.33%
Answer:
7/2 = 35/10
3/5 = 6/10
Step-by-step explanation:
To get a common denominator of 10, simply multiply the current denominator by any number to equal 10. The first fraction has a denominator of 2. If we divide 10 by 2, we get the answer 5. This means 2 can be multiplied by 5 to get a denominator of 10. However, what we do to the denominator we must also do to the numerator (multiply by 5).
7/2
7 × 5 / 2 × 5
35/10
3/5
3 × 2 / 5 × 2
6/10