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melisa1 [442]
3 years ago
13

WHOEVER ANSWERS FIRST WITH A SERIOUS QUESTION WILL GET BRAINLIEST

Mathematics
1 answer:
shusha [124]3 years ago
6 0

Answer: See Explanation

Step-by-step explanation:

Take the amount of money and divide by the ounces for how much it costs per ounce

1.35 / 12 = 0.1125 per ounce

2.10 / 20 = 0.105 per ounce

3.60 / 32 = 0.1125 per ounce

The Big Swig would be the best deal because the cost per ounce is the lowest, getting you the cheaper price for a good size. I believe I did this right, hope I could help.

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Find the cube root of -8.
Lina20 [59]

Answer:

The cube root of -8 is -2

Step-by-step explanation:

Factor the numbers 8=2^3

-∛2³

Apply radical rule;

∛2³=2

=-2

HOPE THIS HELPED!!!

5 0
2 years ago
3x+2y=-4<br><br>I need help solving this. Pease help
marta [7]
Steps to solve:
3x+2y=-4
3x=-4-2y
x=- \frac{4}{3}- \frac{2}{3}y [/tex]
6 0
3 years ago
Gizela teaches 5 yoga classes that are each 3/4 hour. Use the drop down menu to complete the equation below to find the number o
Assoli18 [71]

Answer:

3.75 hours (3 hours and 45 minutes)

Step-by-step explanation:

5 times 3/4 = 3.75

4 0
3 years ago
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
Marrrta [24]

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.

3 0
3 years ago
Please help and add steps!!
Veseljchak [2.6K]

Answer:

so what yhu want to do is 9 times 3  than 5y times 37

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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