Complete question:
The number of concurrent users of a social networking site has increased dramatically since 2005. By 2014, this social networking site could connect concurrently 70 million users online. The function P(t) = 2.459(1.475)^t, where t is the numbe of years after 2005, models this increase in millions of users. Estimate the number of users of this site that could be onlin concurrently in 2006, in 2010, and in 2013. Round to the nearest million users
Answer:
Kindly check explanation
Step-by-step explanation:
Given the function :
The function P(t) = 2.459(1.475)^t, where t is the numbe of years after 2005
Estimate the number of users of this site that could be online concurrently in:
2006:
t = (2006 - 2005) = 1
P(t) = 2.459(1.475)^t
P(1) = 2.459(1.475)^1
P = 2.459(1.475)
= 3.627025 = 4 million users (nearest million)
2010 :
t = (2010 - 2005) = 5
P(t) = 2.459(1.475)^t
P(5) = 2.459(1.475)^5
P = 2.459(6.981682607421875)
= 17.1679575 = 17 million users (nearest million)
2013:
t = (2013 - 2005) = 8
P(t) = 2.459(1.475)^t
P(8) = 2.459(1.475)^8
P = 2.459(22.404546753)
= 55.09278 = 55 million users (nearest million)
