Here we will use exponential form, which is

Initial value, y0 = 500
Double of 500 = 1000



Required equation is

Answer:

Step-by-step explanation:
Given: There are 2 classes of 25 students.
13 play basketball
11 play baseball.
4 play neither of sports.
Lets assume basketball as "a" and baseball as "b".
We know, probablity dependent formula; P(a∪b)= P(a)+P(b)-p(a∩b)
As given total number of student is 25
Now, subtituting the values in the formula.
⇒P(a∪b)= 
taking LCD as 25 to solve.
⇒P(a∪b)= 
∴ P(a∪b)= 
Hence, the probability that a student chosen randomly from the class plays both basketball and baseball is
.
Answer:
Your answer will be 6^3x^5
Answer:
Pls copy and paste the anwser so i can see better-sincerly Dillin
Step-by-step explanation: