The answer is 4 m/s.
In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).
The relation of speed (v), distance (d), and time (t) can be expressed as:
v = d/t
We need to calculate the speed of the second 5 seconds of the travel:
d = 20 m (total 30 meters - first 10 meters)
t = 5 s (time from t = 5 seconds to t = 10 seconds)
Thus:
v = 20m / 5s = 4 m/s
Hey I can help give me one moment .
just separate the ramge and domain
A) 
B)In 200 times he can hit 59 times !
<u>Step-by-step explanation:</u>
Here we have , A baseball player got a hit 19 times in his last 64 times at bat. We need to find the following :
a. What is the experimental probability that the player gets a hit in an at bat?
According to question ,
Favorable outcomes = 19
Total outcomes = 64
Probability = (Favorable outcomes)/(Total outcomes) i.e.
⇒ 
⇒ 
b. If the player comes up to bat 200 times in a season, about how many hits is he likely to get?
According to question , In 64 times he hit 19 times . In 1 time there's probability to hit 0.297 times! So ,In 200 times he can hit :
⇒ 
⇒ Hit = 59.36
Therefore , In 200 times he can hit 59 times !