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Annette [7]
3 years ago
10

Is 13/10 irrational or rational

Mathematics
2 answers:
Akimi4 [234]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

Both 13 and 10 are integers, meaning 13.10 is rational.

motikmotik3 years ago
4 0
It is rational 13/10
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Please answer correct with an explanation thanks!
Strike441 [17]

Answer:

17/50

Step-by-step explanation:

P(red or purple) = number of red or purple / total

There are 7 red rings and 10 purple rings

                          = (7+10) / 50

                          = 17/50

3 0
2 years ago
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What is the Definition of Algebra?
saveliy_v [14]

Answer:

the part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations.

Step-by-step explanation:

7 0
3 years ago
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How do i solve this radical expression? It has to do with factoring i know that.
MAVERICK [17]
Firstly it's not a radical expression.

You can factorize the numerator x²+12x+36 = (x+6)²

Factorize the denominator 12x+72 = 12(x+6)


=(x+6)² / 12(x+6) , simplify. Answer = (x+6)/12
5 0
3 years ago
There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one
diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C

P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)

P(D2/D1) = 0.02667

6 0
3 years ago
It took Sergio 1.5 hours to read 3 chapters of a book. If he continues reading at a constant rate and the chapters are similar i
ryzh [129]

Answer:

9 hours

Step-by-step explanation:

1 chapter ≈ .5 hour

18 x .5 = 9

8 0
2 years ago
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