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inessss [21]
3 years ago
12

Need Help Fast!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
AleksandrR [38]3 years ago
7 0
7.2, 3, 8.09, 2.22, 5.06, 2.5
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Lindsey earns $70 for working 5 hours. How much does she earn for working 12 lawns?
chubhunter [2.5K]

How long does it take her to do a lawn? If it took her 5 hours for 1 lawn then it would be $840 for twelve lawns.


She is getting paid $14 an hour.

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3 years ago
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A box of chalk and 2 staplers cost $10. three boxes of chalk and 2 staplers cost $18. find the total cost of 1 box of chalk and
Alchen [17]
Here is the solution of the given problem above:
Let s= staples 
<span>c = chalk</span>

<span>c + 2s = 10 </span>
<span>3c + 2s = 18 </span>

<span>-c-2s = -10 </span>
<span>3c + 2s = 18 </span>

<span>2c = 8 </span>

<span>c = 4 </span>
<span>s = 3 
</span>Therefore, the price per chalk box is $4 and per stapler is $3.
So the total cost of the chalk and stapler would be $7. 
Hope this is the answer that you are looking for. Thanks for posting it here in brainly.
7 0
4 years ago
Which choices correctly describe reflections in the diagram? Check all that apply.
inessss [21]

Answer:

2

3

5

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8 0
4 years ago
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the table shows the amount of apple sause made from one apple of each size. Patrice has 17 medium apples and 23 large apples. Wh
joja [24]

Answer:

where is the table?

Step-by-step explanation:

7 0
4 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
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