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KiRa [710]
3 years ago
8

Simplify

="TexFormula1" title=" \sqrt[6]{5} \times \sqrt[2]{5} = " alt=" \sqrt[6]{5} \times \sqrt[2]{5} = " align="absmiddle" class="latex-formula">
sumplify
​
Mathematics
1 answer:
jek_recluse [69]3 years ago
5 0

Step-by-step explanation:

\sqrt[6]{5}  \times  \sqrt[2]{5}

=  {5}^{  \frac{1}{6}  }  \times  {5}^{ \frac{1}{2} }

=  {5}^{ \frac{1}{6} +  \frac{1}{2}  }

=  {5}^{ \frac{1 + 3}{6} }

=  {5}^{ \frac{4}{6} }

=  {5}^{ \frac{2}{3} }

=  \sqrt[3]{ {5}^{2} } (ans)

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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar
yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are  K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

\frac{270}{455}=0.5934=59.34\%

(b)

The probability of selecting three 40-W bulbs is

\frac{4*3*2}{455}=0.0527=5.27\%

The probability of selecting three 60-W bulbs is

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The probability of selecting three 75-W bulbs is

\frac{6*5*4}{455}=0.2637=26.37\%

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\frac{6*5*4}{455}=0.2637=26.37\%

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The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is \frac{9}{15}=0.6

Since there are no replacement, the probability of taking a second non 75-W bulb is now \frac{8}{14}=0.5714

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0.6, 0.5714, 0.5384, 0.5, 0.4545

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0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

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