Comparing account balances in exercise, $2000 is deposited in an account. decide which account, (a) or (b), will have the greater balance after 10 years.
(a) 6 1/2%, compounded monthly
(b) 6 1/4%, compounded continuously

account b will have the greater balance after 10 years.

Step-by-step explanation:

a) $2000 is deposited in an account. This means that the principal,

P = 2000

It was compounded monthly. This means that it was compounded 12 times in a year. So

n = 12

The rate at which the principal was compounded is 6.5%. So

r = 6.5/100 = 0.065

It was compounded for just 10 years. So

t = 10

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years. Therefore

A = 2000 (1+0.065/12)^12×10

A = $3824

b) The formula for continuously compounded interest is A = P x e (r x t)

Where

e = 2.7183

A = 2000 × 2.7183^(0.0625 × 10)

A = $3736.5

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

$15 each

step-by-step explanation:

$45/3 children= $15 each