Answer:
The correct answer is 25.2 in.
Explanation:
It is given that number line goes from 0 to 60 which can be used to represent a ribbon of length = 60 inches.
2 inches of the ribbon are frayed so actual length = 58 inches
Please refer to the attached image for the ribbon.
A is at 0
C is at 60
B is at 2
P is the point to divide the remaining ribbon in the ratio 2:3.
Part AB of the ribbon is frayed.
BP: PC = 2:3
Let BP = 2
and PC = 3
![x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/p9sq9b3rc5nwoqzhzc8wcaj51b36281l9g.png)
Now, BP + PC = BC = 58 = 2
+ 3
= 5
![x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/p9sq9b3rc5nwoqzhzc8wcaj51b36281l9g.png)
So,
![5x =58\\\Rightarrow x =11.6](https://img.qammunity.org/2021/formulas/mathematics/high-school/n2zq598pxq2y56avmwzjzlwnacaivhqj1w.png)
BP =
![2* x = 2 * 11.6 = 23.2\ inches](https://img.qammunity.org/2021/formulas/mathematics/high-school/4neos9ghbtxp4htpvqpe6csymcuig81e4u.png)
Location of the Cut = 2 + 23.2 = 25.2 inches
Alternatively, we can use the formula directly:
![x = \frac{m} {m + n } (x_2 - x_1) + x_1](https://img.qammunity.org/2021/formulas/mathematics/high-school/1ld7hyw43rd9vhftdqkphzm42lhl7tbpx2.png)
![x_1 = 2\\x_2 = 60](https://img.qammunity.org/2021/formulas/mathematics/high-school/95wxrtgb59bgga9d6a1vfmg00ilpkca6q4.png)
m: n is the ratio 2:3
![x = \frac{2} {2 +3 } (60- 2) + 2\\\Rightarrow 0.4 (58 )+2\\\Rightarrow 23.2+2 \\\Rightarrow \bold{25.2\ inches}](https://img.qammunity.org/2021/formulas/mathematics/high-school/k1l6rvvk814at911xswuvivqqxvfez7gaj.png)