Audrey invested \$620$620 in an account in the year 2000, and the value has been growing exponentially at a constant rate. The value of the account reached \$700$700 in the year 2004. Determine the value of the account, to the nearest dollar, in the year 2011.

$866

Step-by-step explanation:

The exponential expression to use is:

P(t) = P0e^kt

P0 is the initial population

P(t) is the population after time t;

k is the constant

If Audrey invested $620 in an account in the year 2000, then, P0 = 620

If the value of the account reached $700 in the year 2004, this means after 4 years P(t) is $700

Substitute t = 4 and P(t) = 700 into the expression above;

700 = 620e^4k

700/620 = e^4k

1.129 = e^4k

ln 1.129 = ln(e^4k)

0.1214 = 4k

k = 0.1214/4

k = 0.03034

To determine the value of the account, to the nearest dollar, in the year 2011, that is after 11 years

P(11) = 620e^11k

P(11) = 620e^11(0.03034)

P(11)= 620e^0.3337

P(11) = 620(1.3961)

P(11) = 865.597

Hence the value of the account in 2011 is $866 to the nearest dollars

24+x=y

step-by-step explanation:

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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