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Ganezh [65]
2 years ago
13

Consider the two-way table below. A 4-column table has 3 rows. The first column has entries A, B, total. The second column is la

beled C with entries 15, 9, 24. The third column is labeled D with entries 21, 25, 46. The fourth column is labeled Total with entries 36, 34, 70. Find P(B|C). P(B|C) = StartFraction 9 Over 34 EndFraction = 0. 26 P(B|C) = StartFraction 9 Over 24 EndFraction = 0. 38 P(B|C) = StartFraction 9 Over 70 EndFraction = 0. 13 P(B|C) = StartFraction 9 Over 58 EndFraction = 0. 16.
Mathematics
1 answer:
Nat2105 [25]2 years ago
4 0

The probability is the ratio of the favorable event to the total number of events. Then the P(B|C) is 0.38. Then the correct option is B.

<h3>What is probability?</h3>

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen.

According to the condition, the table will be

                      C                   D               Total

  A                 15                  21                 36

  B                  9                  25                34

Total              24                 46                70

The probability of the (BIC) will be

\rm P(B|C) = \dfrac{9}{24}\\\\\\P(B|C) = 0.375 \approx 0.38

The P(B|C) is 0.38. Then the correct option is B.

More about the probability link is given below.

brainly.com/question/795909

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Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

\large\underline{\sf{Solution-}}

Given expression is

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

On substituting directly x = 1, we get,

\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}

\rm \: = \sf \: \: - \infty \: - \: \infty

which is indeterminant form.

Consider again,

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

can be rewritten as

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]

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\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]

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\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}

\rule{190pt}{2pt}

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3 years ago
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