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JulsSmile [24]
4 years ago
6

What is the relation between real number and whole number

Mathematics
1 answer:
cestrela7 [59]4 years ago
6 0
The set of real numbers is all numbers that can be shown on a number line. It also includes rational numbers, which are numbers that can be written as a ratio of two integers, and irrational numbers, which cannot be written as a the ratio of two integers.
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PLEASE HELP ME I AM FAILING. I WILL GIVE BRAINLIEST
den301095 [7]

Answer:

3.

equation = \sqrt{7^2 + 12^2 } = c

Using:

\sqrt{a^2 + b^2 } = c or a^2 + b^2 = c^2 or

7^2 + 12^2 = c^2

49 + 144 = c^2

c^2 = 193

missing length = 13.9 inches

4. equation = \sqrt{24^2 + 32^2} = c

Using:

\sqrt{a^2 + b^2 } = c or a^2 + b^2 = c^2

24^2 + 32^2 = c^2

reduce to:

8 multiplied by everything:

3^2 + 4^2 = c^2

Since we know 3,4,5 is a Pythagorean triple, we can use this to find c.

Since c must be 5, we multiply by 8 since we divided by 8 earlier to get the 3,4,5 and get 40.

missing length = 40 inches

3 0
3 years ago
3/4+5/36+7/144+.......+17/5184+19/8100<br><br>(a)0.95 (b)1 (c) 0.99 (d)0.98​
abruzzese [7]

Answer:

B) 1

Step-by-step explanation:

All I did was simplify the equation

1509/1600

(Decimal: 0.943125)

HOPE THIS HELPS!!!! :)

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4 0
4 years ago
What is the answer to 9c+3+13c–4c–8c ???
Tanzania [10]

Answer:

The answer is 10c+3

8 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
Solve for x: 5(x-2)=2(10 x)
OverLord2011 [107]
Distribute
5x-10=20+2x
minus 2x both sides
3x-10=20
add 10 both sides
3x=30
divide both sides by 3
x=10
6 0
3 years ago
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