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satela [25.4K]
3 years ago
13

Please help me I have been stuck on this question for ages

Mathematics
1 answer:
Finger [1]3 years ago
6 0

Answer:28 green and 35 red

Step-by-step explanation:

Given

If there are r red counter and g green counter then

Probability of drawing a green counter is P(g)=\frac{4}{9}

and P(g)=\frac{\text{No of g counter}}{\text{Total no of counter}}

Thus \frac{\text{No of g counter}}{\text{Total no of counter}}=\frac{4}{9}

\frac{g}{g+r}=\frac{4}{9}

\Rightarrow 9g=4g+4r

\Rightarrow 5g=4r\quad \ldots(i)

Also if 4 red and 2 green counter is added the probability of drawing a green counter is

P(g)=\frac{10}{23}=\frac{\text{No of g counter}}{\text{Total no of counter}}

\Rightarrow \frac{10}{23}=\frac{g+2}{g+2+r+4}

\Rightarrow \frac{10}{23}=\frac{g+2}{g+r+6}

\Rightarrow 10g+10r+60=23g+46

\Rightarrow 10r+14=13g\ quad \ldots(ii)

Substitute the value of g in equation (ii)[/tex]

\Rightarrow 10\times \frac{5}{4}g+14=13g

\Rightarrow \frac{25}{2}g+14=13g

\Rightarrow g=28

Therefore r=35

Thus there 28 green counter and 35 red counter

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Step-by-step explanation:

We can let r and b represent the numbers of teams in the red and blue divisions, respectively. The total number of goals scored in each division will be the average for that division times the number of teams in that division.

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Solving by substitution, we have ...

  r = b +1

  4.5(b +1) +4.2b = 48 . . . . substitute for r

  8.7b +4.5 = 48 . . . . . . . . simplify

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  r = b +1 = 6 . . . . . . . . . find r

There are 6 red teams and 5 blue teams.

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<em>Additional comment</em>

The basic idea is that you make an equation for each relation given in the problem statement. For a problem like this, you do need to have an understanding of how the average number of goals would be calculated and how that relates to the total goals.

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see the attached figure with the letters

1) find m(x) in the interval A,B
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2) find m(x) in the interval B,C
B(50,40)  C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
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3) find n(x) in the interval A,B
A (0,0)  B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
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4) find n(x) in the interval B,C
B(50,60)  C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
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5) find h(x) = n(m(x)) in the interval A,B
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h(x)=(-36/25)x+120
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6) find h(x) = n(m(x)) in the interval B,C
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then 
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for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

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