Remark
The short answer is you multiply 0.6 times the cm/s to get m/min.
Solve
Though you didn't ask for it, here's the way it is done. Notice that each set of brackets cancels the units of a set of brackets to the left of the set of brackets you are observing. This is called unit analysis. The answer is given below.
![\frac{18 cm}{sec} *[\frac{60 cm}{1 min}] * [\frac{1 m}{100 cm}] = 18*0.6\frac{m}{min}=10.8\frac{m}{min}](https://tex.z-dn.net/?f=%20%5Cfrac%7B18%20cm%7D%7Bsec%7D%20%2A%5B%5Cfrac%7B60%20cm%7D%7B1%20min%7D%5D%20%2A%20%5B%5Cfrac%7B1%20m%7D%7B100%20cm%7D%5D%20%3D%2018%2A0.6%5Cfrac%7Bm%7D%7Bmin%7D%3D10.8%5Cfrac%7Bm%7D%7Bmin%7D)
1,360,720,000/24,151,500=56.34%(rounded)
A/B - 90° | C - 42° | D - 48 | E - 132
Step-by-step explanation:
We are told that for a large data set of a average student grades, the third quartile Q3 is found 78.5.
Since we know that third quartile represents the number such that 75% of the data is less than this number. Third quartile is the middle value between median and the highest value of a data set. Third quartile is also known as upper quartile. Third quartile splits lower 75% data from highest 25% of data.
Therefore, third quartile represents that 75% of student grades are less than 78.5 and 25% of students grades are greater than 78.5.
-2* F
You do +3 +2 -4 -1 which equals to 0. Then you do 0 plus -2 which equals -2*F.