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Nookie1986 [14]
2 years ago
12

Q3. A bag contains 7 red balls and 3 green balls. A ball is taken out replaced.

Mathematics
1 answer:
ycow [4]2 years ago
7 0

Part (a)

There are 7 red out of 7+3 = 10 total

<h3>Answer: 7/10</h3>

==========================================================

Part (b)

We have 3 green out of 10 total

<h3>Answer: 3/10</h3>

==========================================================

Part (c)

3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)

So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.

<h3>Answer: 9/100</h3>

==========================================================

Part (d)

Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.

(7/10)*(7/10) = 49/100

<h3>Answer: 49/100</h3>

==========================================================

Part (e)

7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.

(7/10)*(3/10) = 21/100

<h3>Answer: 21/100</h3>

==========================================================

Part (f)

We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).

<h3>Answer: 21/100</h3>
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4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

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\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

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