Spartan credit bank is offering 7.5 percent compounded daily on its savings accounts. you deposit $5,900 today. a. how much will you have in the account in 4 years? (use 365 days a year. do not round intermediate calculations and round your answer to 2 decimal places, e. g., 32.16.) b. how much will you have in the account in 12 years? (use 365 days a year. do not round intermediate calculations and round your answer to 2 decimal places, e. g., 32.16.) c. how much will you have in the account in 19 years?

1.78 years

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Fourth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

There are four half lives. So, time has passed = 4*162.7 days = 650.8 days = 1.78 years

650.8 days

Explanation:

if the half-life is 162.7 days and the number of half-life days has elapsed 4 times.

To know the number of days that have passed we must multiply the half-life by 4

A) answer 1.281 grams

B) 52.60 days

C) 23250 years  

D)  864 years  

Explanation:

A) Solution

Ending Amount = Beginning Amount / 2(time / half-life)

Ending Amount = 164 / 2(⁴⁴⁸/⁶⁴)

Ending Amount = 164 / 2⁷

Ending Amount = 164 / 128

Ending Amount = 1.281 grams

B) solution

Half life = (time * log 2) / log (beginning amount / ending amount)

Half life = (263 * .30103) / log (5.12 / 0.16)

Half life = 79.17 / log (32)

Half life = 79.17 / 1.505

Half life =  52.60 days

C) Solution

As we have half-life =5730  

1/16 =  6.2 percent  

Remains =    23250 years  

D) solution

elapsed time = half life * log (beginning amount / ending amount) / log 2

elapsed time = 432 * (log (0.29 / 0.0725) / .30103)

elapsed time = 432 * (0.602 / .30103)

elapsed time=  432 (2)

= 864 years  

We are given:

Money on the fifth day (a₅) = $80

Money on day 21 (a₂₁) = $160

To Find:

Money in the bank on the first day (a₁) = ?

Money being added to the account per day (d) = ?

Money in the bank on day 30 (a₃₀) = ?

Finding the money on day one and the money added:

Solving for equations

Since we are adding a constant amount to our account every day, the amount of money in the bank can be represented by an AP

We know that the formula for the nth term of an AP is:

aₙ = a₁ + (n-1)d

So, we can write the amount of money on day 5 as:

a₅ = a₁ + (5-1)d

a₅ = a₁ + 4d

80 = a₁+  4d                        [We are given that a₅ = 80]

We are also given the amount of money on day 21:

a₂₁ = $160

Money on day 21 can be rewritten as:

a₂₁ = a₁ + (21-1)d

160 = a₁ + 20d                  [Since a₂₁ = 160]

Solving the equations:

Now we have 2 equations. We will now solve them to get the values of a₁ and d

160 = a₁ + 20d (1)

80 = a₁ + 4d(2)

From equation (2), we can say that:

a₁ = 80 - 4d              [subtracting 4d from both sides]

Now, we will use this value of a₁ in the equation (1):

160 = a₁ + 20d

160 = (80-4d) + 20d

160 = 80 + 16d

80 = 16d             [subtracting 80 from both sides]

d = 5                   [dividing both sides by 16]

Now that we know the value of d, we can use it in any of the 2 equations:

80 = a₁ + 4d

80 = a₁ + 4(5)              [since d=5]

80 = a₁ + 20

a₁ = 60                        [subtracting 20 from both sides]

Now, we know the values of the initial amount of money and the money added to the account everyday

Initial money in the bank = $60

Money added everyday = $5

Money in day 30:

We will use the AP formula to find the value of money on day 30

a₃₀ = a₁ + (30-1)d

We know that a₁ = $60 , d = $5

a₃₀ = 60 + 29(5)

a₃₀ = 60 + 145

a₃₀ = $205

Money on day 30 = $205