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Maksim231197 [3]
3 years ago
13

A bag contains only red and blue marbles. Yasmine takes one marble at random from the bag. The probability that she takes a red

marble is 1 in 5. Yasmine returns the marble to the bag and adds five more red marbles to the bag. The probability that she takes one red marble at random is now 1 in 3. How many red marbles were originally in the bag?
answer it and get 98 points if you get it right
Mathematics
1 answer:
mamaluj [8]3 years ago
5 0

Let red marbles = X.

The probability is 1 out of 5, written as 1/5

1/5 in terms of red marbles is equal to the number of red marbles divided by 5x, where 5x is the total number of marbles.

1/5 = x/5x 

Now you have 5x total marbles, x red and 4x blue. 

Add 5 more red and the new probability is: 

(x+5)/(5x+5) = 1/3 

Simplify:

3x+15 = 5x+5 

Now solve for x:

Subtract 3x from both sides:

15 = 2x +5

Subtract 5 from each side:

2x = 10 

Divide both sides by 2:

x = 10/2

X = 5

There were originally 5 red marbles.

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If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?​
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Step-by-step explanation:

\bf➤ \underline{Given-} \\

\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\

\bf➤ \underline{To\: find-} \\

\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\

\bf ➤\underline{Solution-} \\

<u>Let us assume that:</u>

\rm: \longmapsto u =  {x}^{2013}

<u>Therefore, the equation becomes:</u>

\rm: \longmapsto u +  \dfrac{1}{u}  = 2

\rm: \longmapsto \dfrac{  {u}^{2} + 1}{u}  = 2

\rm: \longmapsto{u}^{2} + 1 = 2u

\rm: \longmapsto{u}^{2} - 2u + 1 =0

\rm: \longmapsto  {(u - 1)}^{2} =0

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<u>Now substitute the value of u. We get:</u>

\rm: \longmapsto {x}^{2013}  = 1

\rm: \longmapsto x = 1

<u>Therefore:</u>

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\rm: \longmapsto {x}^{2022}  +  \dfrac{1}{ {x}^{2022} }  = 2

★ <u>Which is our required answer.</u>

\textsf{\large{\underline{More To Know}:}}

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

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