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erica [24]
3 years ago
6

There are $15$ gymnasts in the Olympic women's competition, including $4$ Americans. The gold medal goes to first place, silver

to second, and bronze to third. In how many ways can the medals be awarded to three of the $15$ gymnasts, if exactly one of the Americans wins a medal?
Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

There are 220 ways by which the medals can be awarded to three of the 15 gymnast, if exactly one of the Americans wins a medal

Step-by-step explanation:

From the question, we have;

The number of gymnast in the Olympic women's competition = 15

The number of the gymnast who are Americans = 4

The number of medals awarded = 3 medals

The number of ways hat the medals can be awarded to the three of the gymnast if exactly one of the Americans wins a medal is given as follows;

The number of ways one of the medals can be won by one of the four Americans = ₄C₁ = 4 ways

The number of ways the other two medals can be won by the remaining 11 gymnast = ₁₁C₂ = 55 ways

Therefore, the total number of ways, 'N', the medals can be awarded to three of the 15 gymnast, if exactly one of the Americans wins a medal is given as follows;

N = ₄C₁ × ₁₁C₂

∴ N = 4 × 55 = 220

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A: 45x + 30y = 1350
Since we don’t know how many adults and children are in the group, we use x and y

b: x-intercept= 30 y-intercept=45
To find the x-intercept you need to isolate the variable. 45x/45 = x
Then you do the same thing to the other side. 1350/45 = 30
So x=30
Same thing with the y-intercept.
30y/30 = y 1350/30 = 45
y=45 (Not really sure what it means by “what they represent” but I thinks it’s that there are 30 adult tickets and 45 children tickets )

c: so our points are (30,0) and (0,45) so you would graph that.
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Hope this helps

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Step-by-step explanation:

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