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lora16 [44]
4 years ago
12

The probability of drawing two red candies without replacement is 1335 , and the probability of drawing one red candy is 25 . Wh

at is the probability of drawing a second red candy, given that the first candy is red?
27/35
13/70
13/14
26/175
Mathematics
2 answers:
FrozenT [24]4 years ago
7 0

Answer:

13/14 I think

Step-by-step explanation:


Luba_88 [7]4 years ago
4 0

Answer:

The correct option is C. 13/14

Step-by-step explanation:

\text{Probability of drawing the first candy red = }\frac{2}{5}

Let the probability of drawing second candy red be x

\text{Now, Probability of drawing the two candies red = }\frac{13}{35}

So, Probability of drawing two candies red = Probability of drawing one candy red × Probability of drawing second candy red

\frac{13}{35}=\frac{2}{5}\times x

\implies x = \frac{13}{35}\times \frac{5}{2}

\implies x = \frac{13}{14}

\textbf{The probability of drawing second candy red = }\bf\frac{13}{14}

Therefore, The correct option is C. 13/14

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IRINA_888 [86]

Answer:

<em>Answer is option</em><em> </em><em>b</em><em>)</em><em> </em>

<em>{(x - 3)}^{2}  +  {(y - 4)}^{2}  = 400</em>

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Step-by-step explanation:

the \: question \: states \: that \:  \\ distance \: between \: two \: points \: is \: 20 \: units \:  \\ and \: the \: two \: points \: are \: (x,y) \:  \: and \:  \: (3 ,- 4) \\ distance \: formula \\  =   \sqrt{ {(x1 - x2) }^{2} +  {(y1 - y2)}^{2}  }  \\ on \: substituting \: the \: values \: in \: formula \\ 20 =  \sqrt{ {(x - 3)}^{2} +  {(y - ( - 4))}^{2}  }  \\ 20 =  \sqrt{ {(x - 3)}^{2} +  {(y + 4)}^{2}  }  \\ now \: squaring \: on \: both \: sides \\ 400 =  {(x - 3)}^{2}  +  {(y + 4)}^{2}  \\ hence \: the \: answer \: is \: option \: b)

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

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