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allsm [11]
2 years ago
13

6^9+6^8)/(6^8*6^-9)HELP PLS​

Mathematics
2 answers:
podryga [215]2 years ago
5 0

\huge \: \bf \: Question:—

\sf \longmapsto6^9+6^8)/(6^8 \times 6^{-9})

\huge\bf Solution:—

\sf \longmapsto \:  \dfrac{ {6}^{9} +  {6}^{8}  }{ {6}^{8}  \times  {6}^{ - 9} }

\sf \longmapsto \dfrac{10077696 +  {6}^{8} } { {6}^{8}  \times {6}^{ - 9}  }

\sf \longmapsto \dfrac{10077696+1679616}{ {6}^{8} \times  {6}^{ - 9}  }

\sf \longmapsto \:  \dfrac{11757312}{{6}^{8} \times  {6}^{ - 9}}

\sf \longmapsto \dfrac{11757312}{1679616(6 {}^{ - 9} )}

\boxed{\bf \: As \: a^{-n} = 1/a(n) \: ,}

\sf \longmapsto \dfrac{11757312}{1679616  \bigg(\dfrac{1}{10077696} \bigg )}

\sf \longmapsto \dfrac{11757312}{ \dfrac{1}{6} }

\sf \longmapsto \:  \: 70543872

\boxed{\bf \:  \: \: Convert\: 70543872 \: to\: scientific \: notation}

\sf \longmapsto \: 7.0543872 × 10 {}^{7}

\boxed{ \bf \: \: Convert\: 70543872 \: to\: exponent}

\sf\longmapsto {6}^{9}   \times 7

_________________________________

\boxed{\bf \: Answer\: in\: Number}:—

\boxed{\bf \: 70543872}

\boxed{ \bf \: Answer\: in\: Exponent:—}

\boxed{\bf \: {6}^{9}   \times 7}

\boxed{\bf Answer\: in\: Scientific\: Notation:—}

\boxed{ \bf \: 7.0543872 × 10 {}^{7} }

suter [353]2 years ago
3 0

Step-by-step explanation:

<h3><u>Question</u><u>:</u><u>-</u></h3>
  • To simplify the expression

<h3><u>Expression</u><u>:</u><u>-</u></h3>
  • \frac{( {6}^{9}  +  {6}^{8} )}{( {6}^{8}  \times  {6}^{ - 9} )}
<h3><u>Solution</u><u> </u><u>:</u><u>-</u></h3>

= \frac{( {6}^{9}  +  {6}^{8} )}{( {6}^{8}  \times  {6}^{ - 9} )}

  • <em>[</em><em>In</em><em> </em><em>numerator</em><em> </em><em>we</em><em> </em><em>take</em><em> </em><em>6</em><em>^</em><em>8</em><em> </em><em>common</em><em>]</em>

= \frac{ {6}^{8} ( {6} +  1)}{( {6}^{8}  \times  {6}^{ - 9} )}

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

=  \frac{( {6}^{8} \times 7)}{( {6}^{8}  \times  {6}^{ - 9} )}

  • <em>[</em><em>In</em><em> </em><em>denominator</em><em> </em><em>as</em><em> </em><em>we</em><em> </em><em>know</em><em>,</em><em> </em><em>a</em><em>^</em><em>x</em><em> </em><em>×</em><em> </em><em>a</em><em>^</em><em>y</em><em> </em><em>=</em><em> </em><em>a</em><em>^</em><em>(</em><em>x</em><em>+</em><em>y</em><em>)</em><em>]</em>

= \frac{( {6}^{8}   \times 7)}{( {6}^{8 + ( -9)} )}

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

= \frac{( {6}^{8}   \times 7)}{( {6}^{ - 1} )}

  • <em>[</em><em>As</em><em> </em><em>we</em><em> </em><em>know</em><em>,</em><em> </em><em>1</em><em>/</em><em>a</em><em>^</em><em>-</em><em>1</em><em> </em><em>=</em><em> </em><em>a</em><em> </em><em>]</em>

=  {6}^{8}  \times 7 \times 6

  • <em>[</em><em>On</em><em> </em><em>further</em><em> </em><em>multiplying</em><em> </em><em>of</em><em> </em><em>6</em><em>]</em>

=  {6}^{9}  \times 7

<h3><u>Result</u><u>:</u><u>-</u></h3>

<em><u>In</u></em><em><u> </u></em><em><u>exponential</u></em><em><u> </u></em><em><u>form</u></em><em><u>:</u></em>

=  {6}^{9}  \times  {7}^{1} (ans)

<em><u>In</u></em><em><u> </u></em><em><u>n</u></em><em><u>umerical</u></em><em><u> </u></em><em><u>form</u></em><em><u>:</u></em>

70,543,872 (Ans)

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0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

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To solve this question, we need to use the binomial and the normal probability distributions.

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Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that \mu = 100, \sigma = 15

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

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Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

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