Lamar is considering two loans. loan a loan b principal $4,500 $4,300 interest rate 5% 6% time in years 3 3 which loan will have the lowest total payback? loan a will have the lowest total payback because the amount of interest paid is less. loan a will have the lowest total payback because the total of the principal and the interest paid for loan a is less than the same total for loan
b. loan b will have the lowest total payback because the interest is less than the interest for loan a loan b will have the lowest total payback because the total of the principal and the interest paid for loan b is less than the same total for loan
a.

Loan B will have the lowest total payback because the interest is less than the interest for loan A

Step-by-step explanation:

We have given that :

For loan A                                                 For loan B

Principal(PV) =$4500                                Principal(PV)=$4300

Interest rate(r)= 5%                                    Interest rate(r) =6%

Time(n)= 3 years                                        Time(n)= 3 years

Formula used = loan payment =  

when we put values we get,

loan payment of A= $1652.44  and  B=$1608.67

As we see Loan B will have the lowest total payback because the interest is less than A due to which loan payment of B is lowest.

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).