A catering company is catering a large wedding reception. The host of the reception has asked the company to spend a total of $454 on two types of meat: chicken and beef. The

chicken costs $5 per pound, and the beef costs $ 7 per pound. If the catering company

buys 25 pounds of chicken, how many pounds of beef can they buy?

[tex]\frac{10!}{(10-5)!}[/tex]

Step-by-step explanation:

Fiona will attend the restaurant for 5 days, so she will get to choose only 5 out of the 10 menus. At first, recall that the number of ways in which you can pick k elements out of n, without minding the order is [tex]\binom{n}{k} = \frac{n!}{(n-k)!k!}[/tex].

Now, consider that for each group of k elements that you choose, you can order it in k! different ways. Suppose that you have k boxes, that you want to fill out with the k elements you pick. So, for the first box you have k options, for the second one you have k-1 options. Continuing in this fashion, you will have k! differnt ways of ordering the k elements.

So, in total, when order matters, you have [tex]\binom{n}{k}\cdot k![/tex] ways of choosing.

For our case, we have n=10 and k=5, which gives us

[tex]\binom{10}{5}\cdot 5! = \frac{10!}{(10-5)! 5!}\cdot 5! = \frac{10!}{(10-5)!}[/tex]

8 chicken, 4 beef

Step-by-step explanation:

No. of chicken: x

No. of beef: y

x + y = 12

x = 12 - y

10x + 14y = 136

10(12 - y) + 14y = 136

120 - 10y + 14y = 136

4y = 16

y = 4

x = 12 - 4

x = 8

The answer is 47 pounds

Explanation:

1. First, let's calculate the amount of money that was spent on chicken

$5 per pound of chicken x 25 pounds = $125

2. Calculate the amount of money left to buy beef by subtracting the total spend on chicken to the total of the budget.

$454 (total) - $125 (chicken) = $329

3. Calculate how many pounds of beef you can buy with the money left by dividing the money into the price for one pound.

$329 / $7 = 47 pounds

5x 240= 1200

3x 240= 720

Total dinners served at the banquet = 42

Step-by-step explanation:

chicken:beef:fish = 7:5:2

Chicken = [tex]\frac{7}{2} Fish[/tex] = [tex]\frac{7}{5} Beef[/tex]

Beef = [tex]\frac{5}{2} Fish[/tex] = [tex]\frac{5}{7} Chicken[/tex]

Fish = [tex]\frac{2}{7} Chicken[/tex] = [tex]\frac{2}{5} Beef[/tex]

Fish > 5...........(i)

Beef + Fish < 30

[tex]\frac{5}{2} Fish[/tex] + Fish < 30

Fish < 8.57........(ii)

Beef + [tex]\frac{2}{5} Beef[/tex] < 30

Beef < 21.43.........(iii)

[tex]\frac{5}{7} Chicken[/tex] + [tex]\frac{2}{7} Chicken[/tex] < 30

Chicken < 30.........(iv)

Chicken < 25........(v)

[tex]\frac{7}{2} Fish[/tex] < 25

Fish < 7.143........(vi)

[tex]\frac{7}{5} Beef[/tex] < 25

Beef < 17.857........(vii)

From (i), (ii) and (vi)

Fish > 5; Fish < 8.57; Fish <7.143;

Therefore: 5 < Fish < 7.143

So fish is either 6 or 7

From (iii) and (vii)

Beef < 21.43; Beef < 17.857;

Therefore Beef < 17.857

If Fish = 6, Beef = 15;

If Fish = 7, Beef = 17.5

From (iv) and (v)

Chicken < 30; Chicken <25;

Therefore Chicken < 25

If Fish = 6, Chicken = 21;

If Fish = 7, Beef = 24.5

Since meals must be whole numbers, Fish = 6, Beef = 15, Chicken = 21

Total = chicken + beef + fish = 21 + 15 + 6 = 42

To use the graph, we must determine which variable represents pens and which represents pencils. Since we know there are more pencils than pens, and the region graphed goes higher on the y-axis than it does on the x-axis, we determine that y represents the number of pencils and x represents the number of pens.

Now we check each answer choice using the graph.

For A, 300 pens and 900 pencils, we go to 300 on the x-axis and 900 on the y-axis. This is not in the graphed region, so this is not a reasonable solution.

For B, 200 pens and 300 pencils, we go to 200 on the x-axis and 300 on the y-axis. This is on the border of the region, so this is a reasonable solution.

For C, 300 pens and 200 pencils, we go to 300 on the x-axis and 200 on the y-axis; this is not in the graphed region, so this is not a reasonable solution.

For D, 100 pens and 200 pencils, we go to 100 on the x-axis and 200 on the y-axis; this is within the graphed region, so this is a reasonable solution; however, the store would likely want to purchase as many as possible, so this is not the best solution.

Step-by-step explanation:

Answer

125 of each meal are ordered

Step-by-step explanation:

You first multiply 7 x 125, which will get you 875

Next, you multiply 5 x 125 and get 625

Add that together and you get 1500

42 dinners

Step-by-step explanation:

The ratio chicken:beef:fish is 7k:5k:2k

More than 5 fish dinners were served. The smallest whole number greater than 5 is 6. At least 6 fish dinners were served.

The ratio of the dinners is 7k:5k:2k

2k >= 6

k >= 3

Using k = 3, we get:

21:15:6, and 21 + 15 + 6 = 42 total meals

Then beef + fish = 21 + 6 = 27 which is less than 30.

15 is less than 25

Both conditions are met.

Using k = 4, we get:

28:20:8, and 28 + 20 + 8 = 56 total meals

Then beef + fish = 28 + 8 = 36 which is greater than 30.

The first condition is not met, so this cannot be the answer.

42 dinners

Step-by-step explanation:

So the minimum amount of dinners served was 42.

You can download the answer here

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