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Aneli [31]
2 years ago
5

How do I solve the length of AC?​

Mathematics
1 answer:
Crank2 years ago
7 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

∆ABC and ∆ADE are similar triangles.

Thus ;

\frac{AD}{AC}  =  \frac{DE}{BC}  =  \frac{AE}{AB}  \\

So :

\frac{5}{3 + 2x}  =  \frac{4}{8}  \\

\frac{5}{3 + 2x}  =  \frac{1}{2}  \\

Inverse both sides

\frac{3 + 2x}{5}  =  \frac{2}{1}  \\

\frac{2x + 3}{5}  = 2 \\

Multiply sides by 5

5 \times  \frac{2x + 3}{5}  = 5 \times 2 \\

2x + 3 = 10

Thus ;

AC = 2x + 3 = 10

AC = 10

Done...

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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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>

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Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

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There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

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(d)

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

Following this procedure 5 times, we find the probabilities

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As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

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

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