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Elza [17]
2 years ago
6

The value of

Mathematics
1 answer:
marusya05 [52]2 years ago
4 0

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

We have to find out the value of the fraction.

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

\sf \longmapsto x =2 +   \dfrac{1}{2 +  \dfrac{1}{2 +  \dfrac{1}{2 + ... \infty} } }

<u>We can also write it as:</u>

\sf \longmapsto x =2 + \dfrac{1}{x}

\sf \longmapsto x =\dfrac{2x + 1}{x}

\sf \longmapsto  {x}^{2}  =2x + 1

\sf \longmapsto {x}^{2}  - 2x - 1 = 0

<u>Comparing </u>the given <u>equation</u> with <u>ax² + bx + c = 0,</u> we get:

\sf \longmapsto\begin{cases} \sf a =1 \\ \sf b =  - 2 \\ \sf c =  - 1 \end{cases}

<u>By quadratic formula:</u>

\sf \longmapsto x =  \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{8} }{2}

\sf \longmapsto x =  \dfrac{2 \pm2 \sqrt{2} }{2}

\sf \longmapsto x = 1 \pm\sqrt{2}

\sf \longmapsto x = \begin{cases} \sf 1  + \sqrt{2} \\ \sf 1 -  \sqrt{2}  \end{cases}

<u>But </u><u>"</u><u>x"</u><u> cannot be negative. Therefore:</u>

\sf :\implies x = 1 + \sqrt{2}

So, the value of the fraction is 1 + √2.

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a and b represent the values within you'll find 3% of the population.

Until a you'll find 0.485 of the population and until b you'll find 0.485+0.03=0.515

To calculate both unknown values you have to use the standard normal distribution. This distribution is centered in zero, the left tail is negative and the right tail is positive.

a and b are at a equal distance from the mean but with different sign, so you only have to calculate on and then you invert the sign to get the other one.

I'll calculate the positive value:

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Look in the body of the Z-table, rigth or positive entry and reach the margins for the value:

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Then using the values of the mean and the standard deviation you can calculate the corresponding valueof platelet count:

b= (X- μ )/σ

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