Mathematics, 17.10.2020 14:01, mparra4761

On 1st January 2020, Laurie invests P dollars in an account that pays a nominal annual interest rate of 5.5%, compounded quarterly. The amount of money in Laurie’s account at the end of each year follows a geometric sequence with a common ratio, r. Find the value of r. Also, Laurie makes no further deposits to or withdrawals from the account. Find the year in which the amount of money in Laurie’s account will become double the amount she invested.

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answered: emhunter101

Using compound interest and a geometric sequence, it is found that:

The common ratio is q = 1.0561

.The amount of money will double in the year 2032.

Compound interest:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

A(t) is the amount of money after t years. P is the principal(the initial sum of money). r is the interest rate(as a decimal value). n is the number of times that interest is compounded per year. t is the time in years for which the money is invested or borrowed.

In this problem:

Rate of 5.5%, hence r = 0.055

.Compounded quarterly, hence n = 4

Hence, considering the end of each year, that is, t = 1 , the common ratio will be:

$q = \left(1 + \frac{r}{n}\right)^{n}$

$q = \left(1 + \frac{0.055}{4}\right)^{4}$

q = 1.0561

The nth term of a geometric sequence is given by:

$A(n) = A(0)q^{n}$

In which A(0) is the initial value.

In this problem, q = 1.0561 , and the time to double is t for which A(n) = 2A(0) , hence:

$A(n) = A(0)1.0561^{n}$

$2A(0) = A(0)1.0561^{n}$

1.0561^n = 2

$\log{(1.0561^n)} = \log{2}$

$n\log{1.0561} = \log{2}$

$n = \frac{\log{2}}{\log{1.0561}}$

n = 12.69

2020 + 12.69 = 20.32.69.

Hence, the amount of money will double in the year 2032.

A similar problem is given at link

answered: book0001

1) The common ratio = 1.055

2) The year in which the amount of money in Laurie's account will become double is the year 2032

Step-by-step explanation:

1) The given information are;

The date Laurie made the investment = 1st, January, 2020

The annual interest rate of the investment = 5.5%

Type of interest rate = Compound interest

Therefore, we have;

The value, amount, of the investment after a given number of year, given as follows;

Amount in her account = a, a × (1 + i), a × (1 + i)², a × (1 + i)³, a × (1 + i)ⁿ

Which is in the form of the sum of a geometric progression, Sₙ given as follows;

Sₙ = a + a × r + a × r² + a × r³ + ... + a × rⁿ

Where;

n = The number of years

Therefore, the common ratio = 1 + i = r = 1 + 0.055 = 1.055

The common ratio = 1.055

2) When the money doubles, we have;

2·a = a × rⁿ = a × 1.055ⁿ

2·a = a × 1.055ⁿ

2·a/a = 2 = 1.055ⁿ

2 = 1.055ⁿ

Taking log of both sides gives;

㏒2 = ㏒(1.055ⁿ) = n × ㏒(1.055)

㏒2 = n × ㏒(1.055)

n = ㏒2/(㏒(1.055)) ≈ 12.95

The number of years it will take for the amount of money in Laurie's account to double = n = 12.95 years

Therefore, the year in which the amount of money in Laurie's account will become double = 2020 + 12..95 = 2032.95 which is the year 2032

The year in which the amount of money in Laurie's account will become double = year 2032.

answered: Guest

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !
Find the slope of the line parallel to each given line

Find the slope of the line parallel to each given line

answered: Guest

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

i ran through it by the other answer is wrong

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On 1st January 2020, Laurie invests P dollars in an account that pays a nominal annual interest rate...