Answer: Chris reached Boston at 10:24 a.m.
Explanation:
Since we have given that
Speed at which Chris ran from Newton to Boston = 15 km/h
Speed at which Goffy ran from Newton to Boston = 12 km/h
Time at which both started = 10:00 a.m.
Time at which Goffy reached Boston = 10:30 a.m.
That means Goffy takes half an hour to reach Boston,
so, Distance between Newton and Boston is given by
![Speed \times Time\\\\=12\times \frac{1}{2}=6 \ km](https://tex.z-dn.net/?f=Speed%20%5Ctimes%20Time%5C%5C%5C%5C%3D12%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%3D6%20%5C%20km)
So, Time taken by Chris to ran from Newton and Boston is given by
![Time=\frac{Distance}{Speed}=\frac{6}{15}=\frac{2}{5}\times 60\ minutes=24\ minutes\\](https://tex.z-dn.net/?f=Time%3D%5Cfrac%7BDistance%7D%7BSpeed%7D%3D%5Cfrac%7B6%7D%7B15%7D%3D%5Cfrac%7B2%7D%7B5%7D%5Ctimes%2060%5C%20minutes%3D24%5C%20minutes%5C%5C)
So, Chris reached Boston at 10:24 a.m.
Standard: 203
Expanded: 200+3
Answer:
3cm I think that's the answer
Answer:
When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour
Step-by-step explanation:
Ex: If you have a multiple-unit rate such as 120 students for every 3 buses, and want to find the single-unit rate, write a ratio equal to the multiple-unit rate with 1 as the second term.
120/3 = 40/1
The unit rate of 120 students for every 3 buses is 40 students per bus. You could also find the unit rate by dividing the first term of the ratio by the second term.
Not a 100% positive this is the answer but i think it is 2.5
got this by dividing 30 by 12 and got 2.5