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julsineya [31]
3 years ago
13

How do you find the percentage of 3/9??? Please show work

Mathematics
1 answer:
tekilochka [14]3 years ago
4 0
All you do
Now, if you divide 1 (the numerator or dividend) by 3 (the denominator or divisor), you get the quotient 0.3333333333333333. Moving the decimal point 2 places to the right (multiplying by 100) and rounding to two decimal points, we end up with 33.33%.
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The area of a floor to be carpeted is 2700 square feet what is this area in square yards?
Cerrena [4.2K]
C 1yd^2 = 9ft^2 _____yd^2 = 2700ft^2 2700 / 9 = 300 yd^2
5 0
3 years ago
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Use the area model to find the product of 28×32
Rudiy27
When u times 28 by 32 you will get 896..I'm not sure if this is what u were asking for but if so here you go :)
4 0
3 years ago
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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
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
11. Ms. Green tells you that a right triangle has a hypotenuse of 13 and a leg
Bad White [126]

Answer:

12

Step-by-step explanation:

we use Pythagoras law

√13^2-5^2

8 0
3 years ago
The hypotnuse of a 45-45-90 triangle has a length of 10 units whats the length of one of its legs
mamaluj [8]

Answer:

  5√2 units ≈ 7.07 units

Step-by-step explanation:

The ratio of leg to hypotenuse in an isosceles right triangle is ...

  leg/hypotenuse = 1/√2

Multiplying by the length of the hypotenuse, we have ...

  leg = hypotenuse/√2 = 10/√2

  leg = 5√2 . . . . rationalize the denominator

The length of one leg is 5√2 units, about 7.07107 units.

7 0
3 years ago
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