Answer:
Step-by-step explanation:
I'm sure you want your functions to appear as perfectly formed as possible so that others can help you. f(x) = 4(2)x should be written with the " ^ " sign to denote exponentation: f(x) = 4(2)^x
f(b) - f(a)
The formula for "average rate of change" is a.r.c. = --------------
b - a
change in function value
This is equivalent to ---------------------------------------
change in x value
For Section A: x changes from 1 to 2 and the function changes from 4(2)^1 to 4(2)^2: 8 to 16. Thus, "change in function value" is 8 for a 1-unit change in x from 1 to 2. Thus, in this Section, the a.r.c. is:
8
------ = 8 units (Section A)
1
Section B: x changes from 3 to 4, a net change of 1 unit: f(x) changes from
4(2)^3 to 4(2)^4, or 32 to 256, a net change of 224 units. Thus, the a.r.c. is
224 units
----------------- = 224 units (Section B)
1 unit
The a.r.c for Section B is 28 times greater than the a.r.c. for Section A.
This change in outcome is so great because the function f(x) is an exponential function; as x increases in unit steps, the function increases much faster (we say "exponentially").
Wrong because you need to add the -2 to the 8.... I hope this helps
Answer:
Mizuki here to answer! B is the correct answer!
Step-by-step explanation:
1 - 0.4 = 0.6
2.1 x 0.6 = 1.26
Answer: 
Step-by-step explanation:
Given : The fuel efficiency of car = 7.6 km per kilogram
We also given that
1 mile = 1.609 km
Then, 
1 gallon = 3.785 liters

1 liter of gasoline = 0.729 kg
Then, 

Now, the fuel efficiency of car will be :

Hence, the car’s fuel efficiency in miles per gallon= 4.5
Answer:
The question is about the least amount to charge each policyholder as premium
The least premium is $484
Step-by-step explanation:
The least amount of premium to charge for this policy is the sum of the expected values of outcome of both instances of policyholder dying before the age of 70 and living after the age of 70 years
expected value of dying before 70 years=payout*probability=$24,200*2%=$484
Expected of living after 70=payout*probability=$0*98%=$0
sum of expected values=$484+$0=$484
Note that payout is nil if policyholder lives beyond 70 years
The premium of $573 means that a profit of $89 is recorded