MATH SOLVE

2 months ago

Q:
# In problems 16-20, calculate the expected price in the year 2008 if you assume that there was a consistent 5% inflation rate and use the given 1988 price. answers should be rounded to the nearest penny. 16. median salary, $27,225 17. gallon of gas, $1.08 18. dozen eggs, $0.89 19. movie admission, $3.50 20. mcdonalds hamburger, $0.62

Accepted Solution

A:

16. 72236.03

17. 2.87

18. 2.36

19. 9.29

20. 1.65

The exponential growth formula is as follows

[tex]y=a(1+r)^{x}[/tex]:

a is initial value

($27,225 17. gallon of gas, $1.08 18. dozen eggs, $0.89 19. movie admission, $3.50 20. mcdonalds hamburger, $0.62)

r is rate of increase (0.05)

x is time (2008-1988 = 20 years)

to solve plug in values for each one.

[tex]y=27225(1+0.05)^{20} = 72236.03[/tex]

[tex]1.08(1+0.05)^{20} = 2.87[/tex]

[tex].89(1+0.05)^{20} = 2.36[/tex]

[tex]3.50(1+0.05)^{20} = 9.29[/tex]

[tex].62(1+0.05)^{20} = 1.65[/tex]

17. 2.87

18. 2.36

19. 9.29

20. 1.65

The exponential growth formula is as follows

[tex]y=a(1+r)^{x}[/tex]:

a is initial value

($27,225 17. gallon of gas, $1.08 18. dozen eggs, $0.89 19. movie admission, $3.50 20. mcdonalds hamburger, $0.62)

r is rate of increase (0.05)

x is time (2008-1988 = 20 years)

to solve plug in values for each one.

[tex]y=27225(1+0.05)^{20} = 72236.03[/tex]

[tex]1.08(1+0.05)^{20} = 2.87[/tex]

[tex].89(1+0.05)^{20} = 2.36[/tex]

[tex]3.50(1+0.05)^{20} = 9.29[/tex]

[tex].62(1+0.05)^{20} = 1.65[/tex]