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3 years ago

13. A refrigerator has a variety of drinks.

Mathematics

1 answer:

Nonamiya [84]3 years ago

5 0

Answer: D

Step-by-step explanation:

if you where twice as likely to select a water than a cola, they would be flip flopped.

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Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o

Ivahew [28]

Answer:

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

5 0

3 years ago

HELP PLEASE I WILL GIVE THE BRAINLIEST ANSWER Parking meter that is 1.6 m tall cast a shadow 3.6 m long at the same time a tree

Julli [10]

Answer:

I am not able to answer your question because I am unsure of what to solve for. :/

Step-by-step explanation:

4 0

3 years ago

Can someone hep with this question please it seems like i can't get the answer correctly.ty!!

Mashcka [7]

Answer:

your answer will be 5

Step-by-step explanation:

hope it helps have a nice day/night ^_^

bye

4 0

3 years ago

Answer this and get 50 points.

stealth61 [152]

Answer:

I drew more points as uploaded picture.

We have, arc BD + 210 + 90 = 360 deg

(sum of all arcs on circle)

=> arc BD = 360 - 210 - 90 = 60 deg

Here we have angle CBD = (1/2) x arc CD = (1/2) x 90 = 45 deg

(property of angle on circle)

=> angle ABD = angle ABC - angle CBD = 180 - CBD = 180 - 45 = 135 deg

( A, B, C lie on same line => ABC = 180 deg)

Otherwise, we have:

angle BDA = (1/2) x arc BD = (1/2) x 60 = 30 deg

We have: x + angle BDA + angle ABD = 180 deg

(sum of all angles in a triangle)

=> x = 180 - angle BDA - angle ABD

=> x = 180 -135 - 30

=> x = 15 deg

Hope this helps!

7 0

3 years ago

Read 2 more answers

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548 approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: $26\times 26\times 26 \times 10 \times 10 \times10$

$26^3\times 10^3=17576000$

The numbers of ways in which two letters followed by four digits can be chosen is: $26\times 26\times 10 \times 10 \times 10 \times10$

$26^2\times 10^4=6760000$

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: $26^2=676$

The probability that a randomly chosen plate contains the number 2222 is:

$\frac{676}{24336000} \approx 0.000028$

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

$26\times 1\times10^3+1\times 26\times10^3$

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

$1\times 10^4$

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

$1\times 10^4+26\times 1\times10^3+1\times 26\times10^3$

$62000$

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

$\frac{62000}{24336000} \approx 0.002548$

7 0

3 years ago