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qwelly [4]
3 years ago
12

An assistant is in charge of ordering lunch for the monthly office meeting. For the first month, she orders five Falafel wraps,

fifteen Turkey BLT sandwiches, and twenty hot Ham and Swiss paninis for a total of $260. The next month, she orders eight Falafel wraps, eighteen Turkey BLT sandwiches, and 14 hot Ham and Swiss paninis for a total of $263. For the third month, she orders twelve Falafel wraps, 16 Turkey BLT sandwiches, and twelve hot Ham and Swiss paninis for a total of $258. How much does each sandwich cost?
Falafel Wrap= $ ?
Turkey BLT= $ ?
Hot Ham and Swiss= $ ?
Mathematics
2 answers:
gulaghasi [49]3 years ago
7 0

Answer:

The falafel wrap costs $5.50, the turkey BLT costs $7.50, and the ham and swiss panini costs $6.00

Step-by-step explanation:

Let us first write a system of equations to write equations to represent each of the situations. Let the variables were

f=cost of falafel wrap

t=cost of Turkey BLT

h=cost of ham and swiss panini

5f+15t+20h=260

8f+18t+14h=263

12f+16t+12h=258

Next, we will solve these system of equations.

-8/5(5f+15t+20h=260)

multiply equation by -8/5

-8f-24t-32h=-416 multiply

8f+18t+14h=263 add to another equation to remove a variable

-6t-18h=-153

-12/8(8f+18t+14h=263) repeat procedure for another equation

-12f-27t-21h=-394.5 multiply

12f+16t+12h=258 combine with another equation

-11t-9h=-136.5

-2(-11t-9h=-136.5) repeat again by multiplying by -2

22t+18h=273

-6t-18h=-153 combine the two equations

16t=120

t=7.5

Now we can replace the value of t into the previous equation to find the cost of the ham and swiss panini

22(7.5)+18h=273

165+18h=273

18h=108

h=6

Now that we have the cost of two sandwiches, we can find the cost of the falafel wrap by substituting the costs of the sandwiches into the equation.

5f+15(7.5)+20(6)=260

5f+112.5+120=260

5f+232.5=260

5f=27.5

f=5.5

Therefore the falafel wrap costs $5.50, the turkey BLT costs $7.50, and the ham and swiss panini costs $6.00

Andru [333]3 years ago
4 0

Answer:

x (Falafel) = $5.50

y (Turkey BLT) = $7.50

z (Paninis) = $6.00

Step-by-step explanation:

Step 1: Write out systems of equations

5x + 15y + 20z = 260

8x + 18y + 14z = 263

12x + 16y + 12z = 258

There are multiple ways to solve for this systems of equations. I will use an augmented matrix for this:

Top row: [5 15 20 | 260]

Middle row: [8 18 14 | 263}

Bottom row: [12 16 12 | 258]

We find RREF form of the augmented matrix to find our answers:

Top row: [1 0 0 | 11/2]

Middle row: [0 1 0 | 15/2]

Bottom row: [0 0 1 | 6]

And we have our answer!

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A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp
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Step-by-step explanation:

<em>"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."</em>

<em />

<em>"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."</em>

<em>"(i) Find the total number of possible ways that these six pairs can be formed."</em>

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

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<em>"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."</em>

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₆C₂ × ₄C₂ × ₂C₂

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90 × 90 / 7,484,400

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<em>"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.</em>

<em>(i) How many different schedules are possible?"</em>

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There are 6 options for the first match.  After that, there are 5 options for the second match.  Then 4 options for the third match.  So on and so forth.  So the number of permutations is 6!.

The total number of possible schedules is:

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<em>"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."</em>

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

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₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

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If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

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(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

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