Javier invests $500 into an account

with the bank that compounds continuously with an annual rate of 9.0% and

Maria invests $500 into a different account that compounds annually with an

annual rate of 9.03%. Find the value in

each account after 10 and 30 years. What

did you notice? Should you compare more

than just the annual rate to determine

which account to invest in? Why or why

not?

Amount in Javier's Account is $1183.68(after 10 years) and $6633.84(after 30 years)

Amount in Maria's Account is $1186.94 (after 10 years) and $6688.83(after 30 years)

The Interest Rate is Not Sufficient.

Step-by-step explanation:

At Compound Interest

Amount= P(1+r)ⁿ

where:

P=Principal,

R= Rate,

n= number of years

Javier's Principal=$500

Javier's Annual Rate =9%=0.09

I. After 10 years

Amount= P(1+r)ⁿ

= 500(1+0.09)¹⁰=500(1.09)¹⁰ =$1183.68

II. After 30 years

Amount= 500(1+0.09)³⁰=500(1.09)³⁰ =$6633.84

Maria's Principal=$500

Maria's Annual Rate=9.03%=0.0903

I. After 10 years

Amount= P(1+r)ⁿ

= 500(1+0.0903)¹⁰=500(1.0903)¹⁰ =$1186.94

II. After 30 years

Amount= 500(1+0.0903)³⁰ =500(1.0903)³⁰ =$6688.83

The Annual Rate and Time is sufficient enough to help determine which account to invest in. The longer the duration, the more the Amount of money in the Account.

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong