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muminat
3 years ago
6

5th grade Help with Add and subtract fractions with unlike denominators

Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0

Addition of fractions with unlike denominators: 3/10 + 1/6 = 7/15

Subtraction of fractions with unlike denominators: 7/8 - 5/6 = 1/24

<u>Step-by-step explanation</u>:

"Unlike denominator" is said to have different numbers in the denominator.

  • Change the unlike denominators to like denominators by finding the least common multiple (LCM) of the denominators.
  • Now multiply the numerator and denominator with same number that gives the LCM in the denominator.
  • Add or subtract the numerators according to the operator given, but do not perform any operation in the denominators.

<u>Addition of fractions with unlike denominators</u>:

⇒ Add 3/10 and 1/6.

⇒ Taking LCM  2 |<u>10 6</u>

                             <u>|5   3 </u>  

⇒ The LCM is 2\times5\times3 = 30.

⇒ 3/10 + 1/6 = (3\times3)/(10\times3) + (1\times5)/(6\times5)

                     = 9/30 + 5/30

                     = 14/30 = 7/15

<u>Subtraction of fractions with unlike denominators</u>:

⇒ Subtract 7/8 and 5/6.

⇒ Taking LCM  2 <u>|8 6</u>

                             <u>|4  3 </u>  

⇒ The LCM is 2\times4\times3 = 24.

⇒ 7/8 - 5/6 = (7\times3)/(8\times3) + (5\times4)/(6\times4)

                     = 21/24 - 20/24

                     = 1/24

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Step-by-step explanation:

<u>Algebraic Simplification</u>

The rules of algebra can be applied in some situations to make expressions simplified or expanded, according to the specific needs of each problem.

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A box contains 15 resistors. twelve of them are labelled 50ω and the other three are labeled 100ω. what is the probability that
Igoryamba

Answer : P(second resistor is 100ω , given that the first resistor is 50ω) is given by

\frac{1}{5}

Explanation :

Since we have given that

Total number of resistors =15

Number of resistors labelled with 50ω = 12

Number of resistors labelled with 100ω =3

Let A: Event getting resistor with 50ω

B: Event getting resistor with 100ω

Since A and B are independent events .

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P(A\cap B)=P(A).P(B)

Now, According to question , we can get that

P(A)= \frac{12}{15}=\frac{4}{5}\\\\P(B)=\frac{3}{15}=\frac{1}{5}

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P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=\frac{4}{5}\times \frac{1}{5}\\\\P(A\cap B)=\frac{4}{25}

So, by using the conditional probability , which state that

P(B\mid A)=\frac{P(A\cap B)}{P(A)}

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So, P(second resistor is 100ω , given that the first resistor is 50ω) is given by

\frac{1}{5}


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