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guapka [62]
4 years ago
8

(Financial application: compound value) Suppose you save $100 each month into a savings account with the annual interest rate 5%

. Thus, the monthly interest rate is 0.05/12 = 0.00417. After the first month, the value in the account becomes 100 * (1 + 0.00417) = 100.417 After the second month, the value in the account becomes (100 + 100.417) * (1 + 0.00417) = 201.252 After the third month, the value in the account becomes (100 + 201.252) * (1 + 0.00417) = 302.507 and so on. Write a program that prompts the user to enter a monthly saving amount and displays the account value after the sixth month. (In Exercise 5.30, you will use a loop to simplify the code and display the account value for any month.)
Computers and Technology
1 answer:
allochka39001 [22]4 years ago
7 0

Answer:

Follow the code or function below...

Do not forget the saving amount, and initialise the balance with that amount. Inside the loop, work out and add the interest and then add the saving amount for the next month.

Explanation:

inicial = float(eval(input('Enter the montly saving amount: ')))

x = (1 + 0.00417)

month_one = inicial * x

month_two = (inicial + month_one) * x

month_three = (inicial + month_two) * x

month_four = (inicial + month_three) * x

month_five = (inicial + month_four) * x

month_six = (inicial + month_five) * x

print('The sixth month value is: '+str(month_six))

Don't forget the saving amount, and initialise the balance with that amount. Inside the loop, work out and add the interest and then add the saving amount for the next month.

balance = 801

for month in range(6):

   balance = balance * (1.00417)

print(balance)

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The daily power supply and management, coupled with the needed skills and items, can be overwhelming for a company whose main objective is not to deliver database services. So cloud computing is adopted to reduce the cost.

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tatiyna

Answer:

Total Memory= 4 KB = 4096 bytes = 32768 bits

Explanation:

<em><u>1. Data lines are 8 From D0 to D7</u></em>

so

Total memory at single address locations is 8 bits.

<em><u>2. Address lines are 12 (A0 to A11)</u></em>

There are 12 address lines but 3 out 12 are for selction of chip or memory bank.

so only 9 pins are there to address the locations on one chip.

Total No. of address locations on single chip = 2^9 = 512 locations

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<u><em>3. Total Memory Bank </em></u>

There are total 3 selection pins for memory bank.

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Total chips = 2^3 = 8.

<em><u>4. Total Memory </u></em>

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