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dedylja [7]
3 years ago
10

Find the domain of ( (a+b/b) − (a/a+b))÷( (a+b/a) − (b/a+b) )

Mathematics
2 answers:
Alexandra [31]3 years ago
7 0

Answer:

Hence the domain is given as,b is such that b is a member of all real numbers,except b=0,a=0, a=-b

Step-by-step explanation:

The domain refers to values for which the expression is defined.

This implies that, the denominators are not equal to zero.

(\frac{a + b}{b} -  \frac{a}{a + b}) \div ( \frac{a + b}{a} -  \frac{b}{a + b})

(\frac{(a + b)(a + b)-ab}{b(a + b)})\div(\frac{(a + b)(a + b)-ab}{a(a + b)})

\frac{(a + b)(a + b)-ab}{b(a + b)} \times\frac{a(a + b)}{(a + b)(a + b)-ab}

\implies \frac{a}{b}

Hence the domain is given as,b is such that b is a member of all real numbers,except b=0,a=0, and a=-b

lora16 [44]3 years ago
6 0

Answer:a/b

Step-by-step explanation:

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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)

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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

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

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P(A)=1/3

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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

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