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vitfil [10]
3 years ago
7

Arrange the fractions in increasing order 7/15, 2/3, 11/12, 5/9

Mathematics
2 answers:
RoseWind [281]3 years ago
4 0
To compare fraction sizes, you have to make sure the denominators of the fractions are the same. You must find a number that is a multiple of all the denominators we currently have (15, 3, 12, 9)
The smallest number that fits this is 540
Now you must multiply the numerator and denominator of each fraction by the number that makes the denominator 540
So, for 7/15 , we multiply the top and bottom by 36, since you need to multiply 15 by 36 to get 540. This means the fraction you end up with is 252/540
For 2/3, we multiply the top and bottom by 180, since you need to multiply 3 by 180 to get 540.This means the fraction you end up with is 360/540
For 11/12, we multiply the top and bottom by 45, since you need to multiply 12 by 45 to get 540. This means that the fraction you end up with is 495/540
For 5/9, we must multiply the top and bottom by 60, since you need to multiply 9 by 60 to get 540. This means that the fraction you end up with is 300/540

Now that the hard part is out of the way, you have four fractions: 252/540, 360/540, 495/540 and 300/540. Now, you can arrange the fractions in ascending order just based on their numerator. 
So, the order would be: 252/540, 300/540, 360/540, 495/540

Now, for the final step, just convert them back to the original fractions.
So, the final answer is:
7/15, 5/9, 2/3, 11/12

Hope I helped!! xx

Oksana_A [137]3 years ago
4 0
2 ways to do this, the official method which involves finding a common denominator or to do it through mental math

Common denominator.
You want to convert all the fractions so that they have a common denominator, once they have a common denominator, then you can just compare the numerator
If we look at the denominators given, 15,3,12,9, through prime factorization, you can determine that the least common denominator would be 3x3x4x5 which is 120

convert each fraction to have a denominator of 180
7/15 *12/12 = 84/180
2/3 *60/60 =120/180
11/12*15/15 = 165/180
5/9*20/20 = 100/180

increasing order of 84,120,165,100 would be 84,100,120,165
which is equivalent to
7/15 5/9, 2/3,11/12


now, mental math/comparison method
you need to know that 1/2 is less than 2/3 and 2/3 is less than 1

now, 7/15 is very close to 1/2 but slightly lower since 7.5/15=1/2
11/12 is very close to 1 but slightly smaller by 1/12
and 5/9 is very close to 2/3 but slightly smaller, but remember 2/3 is greater than 1/2 so 5/9 is greater than 1/2

so the order based off these comparison would be
7/5,5/9, 2/3, 11/12
because its the equivalent of comparing
1/2,2/3,2/3,1
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Step-by-step explanation:

The area of a rectangle is

A = l*w

70 = (x - 11)(x - 8)

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70 = x^2 -8x-11x+88

Combine like terms

70 = x^2 -19x +88

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0 = x^2 -19x +88-70

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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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