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kirill [66]
3 years ago
13

Rick has taken a loan of $3,200 from the bank to buy new appliances. His first loan payment is due at the end of this month. He

has 24 months to pay off his loan, starting this month. The loan will be compounded monthly at a fixed annual rate of 4.8%.
Use the formula for the sum of a finite geometric sequence to determine Rick's approximate monthly payment.


Rick's approximate monthly payment will be $293.48.

Rick's approximate monthly payment will be $81.27.

Rick's approximate monthly payment will be $128.

Rick's approximate monthly payment will be $140.87.
Chemistry
2 answers:
marysya [2.9K]3 years ago
4 0

Answer:

140

Explanation:

lisov135 [29]3 years ago
3 0

<u>Answer:</u>

Monthly payment of Rick  will be  approximately $ 140.87

<u>Explantion:</u>

GIVEN :

Loan amount P=\$ 3,200

Time duration for loan repayment n=24 \mathrm{months}

Rate of interest  k=4.8 \% \text { per anuum }

To find:

Rick's monthly payment will be  approximately $140.87  

Solution:

Using  sum of a finite geometric sequence:

a=P\left[\frac{[1-r]}{\left[r-r^{n+1}\right]}\right]

where  r=\frac{1}{1+i}

             i=  monthly interest

but we rate of interest per annum K = 4.8%

now , rate of interest per month, i=\frac{4.8}{12}=0.4 \%=\frac{0.4}{100}=0.004

lets find the value of r

r=\frac{1}{1+i}

r=\frac{1}{1.004}        

r=0.995                                                             substituting the values,

a=P\left[\frac{1-r}{r-r^{n+1}}\right]

a=3200\left[\frac{1-0.995}{0.995-0.995^{25}}\right]

a=3200\left[\frac{0.005}{0.995-0.882}\right]

a=3200\left[\frac{0.005}{0.118}\right]

a=3200[0.044]

a=140.8

Result :

Hence the  approximate monthly payment of  Rick  is  $140.87

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Explanation:

Step 1: Data given

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Number of moles nitric acid (HNO3) = 0.25 moles

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Step 2: Calculate molar mass of nitric acid

Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)

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Molar mass HNO3 = 63.01 g/mol

Step 3: Calculate mass of solute use

Mass HNO3 = moles HNO3 * molar mass HNO3

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If you had 240 L container at 479 k and 300 kpa, what would the volume be if you changed the conditions to STP
nydimaria [60]

Answer:

The answer to your question is V2 = 434.7 l

Explanation:

Data

Volume 1 = V1 = 240 l                             Volume 2 = ?

Temperature 1 = T1 = 479°K                   Temperature 2 = T2 = 293°K

Pressure 1 = P1 = 300 KPa                      Pressure 2 = P2 = 101.325 Kpa

Process

1.- Use the combined gas law to solve this problem

                P1V1/T1 = P2V2/t2

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               V2 = P1V1T2 / T1P2

2.- Substitution

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3.- Simplification

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4.- Result

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