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Gennadij [26K]
3 years ago
13

How many pounds of coffee worth $7 a pound should be added to 30 pounds of coffee worth $4 a pound to get a mixture worth $5 a p

ound
Mathematics
2 answers:
EleoNora [17]3 years ago
7 0
It’s a 5 pounds and it’s a long thing so it’s just easy to answer it’s a
goldfiish [28.3K]3 years ago
4 0
Let's begin by assigning a letter to represent our unknown:
x = lbs. of coffee at $7 per lb.

Now let's write an equation to express the cost of mixing the $7 per lb. coffee with the $4 per lb. coffee based upon how many pounds we have of each:
(7)(x) + (4)(10) = (5)(x + 10)

Solving our equation for x:
7x + 40 = 5x + 50
7x - 5x = 50 - 40
2x = 10
x = 5 (lbs. of coffee at $7 per lb.)

We can verify our answer by plugging our answer back into our equation and checking to see that both sides balance:
(7)(x) + (4)(10) = 5(x + 10)
(7)(5) + 40 = 5(5 + 10)
35 + 40 = 25 + 50
75 = 75 (both sides balance)

Since both sides balance, we are confident that our answer is correct.
Hope this helps !
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A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
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Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

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