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ELEN [110]
3 years ago
9

What is thw radius of a circle given by the equation x^2+y^2-2x+8y-47=0

Mathematics
1 answer:
Arlecino [84]3 years ago
8 0

x^2 + y^2 - 2x + 8y - 47 = 0

x^2 + y^2 - 2x + 8y = 47

(x^2 - 2x) + (y^2 + 8y) = 47

(x^2 - 2(1)x) + (y^2 + 2(4)y) = 47

(x^2 - 2(1)x + 1^2) + (y^2 + 2(4)y + 4^2) = 47 + 1^2 + 4^2

(x - 1)^2 + (y + 4)^2 = 64 = 8^2

r=8



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Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

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Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

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Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

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Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

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Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

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probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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