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• ## Related Books

 0 answersPost by Elvis Wodi on February 16, 2015why did the we get one as the result 0 answersPost by Mohamed Kaba on August 11, 2014Instead of using nested "ifs" could we use "&" in the first "if", and then use "else" at the end. For example:if (a~=0) & (d<0)    disp('complex roots')else   x1=....   x2=....end

### Decisions, Part 2

• An if construct can contain further ifs and so on, this action is called nesting, and is different than elseif ladder.
• The switch statement executes certain statements based on the value of a variable or expression.

### Decisions, Part 2

The electricity accounts of residents in a very small town are calculated as follows:
♦  500 units or less are used the cost is 2 cents per unit;
♦  more than 500, but not more than 1000 units are used, the cost is \$10 for the first 500 units, and then 5 cents for every unit in excess of 500;
♦  if more than 1000 units are used, the cost is \$35 or the first 1000 units plus 10 cents for every unit in excess of 1000;
♦  in addition, a basic service fee of \$5 is charged, no matter how much electricity is used.

Write a program which enters the following five consumptions into a vector, and uses a for loop to calculate and display the total charge for each one: 200, 500, 700, 1000, 1500.
cons = [200 500 700 1000 1500];
for i = 1:numel(cons)
if cons(i)  < = 500
cost(i) = cons(i)*0.02 + 5;
elseif cons(i)  < = 1000
cost(i) = 10 + (cons(i) - 500)*0.05 + 5;
else
cost(i) = 35 + (cons(i) - 1000)*0.10 + 5;
end
end
disp('cost = ')
disp(cost)

\$9, \$15, \$25, \$40, \$90
If you invest \$1000 for one year at an interest rate of 12 percent, the return is \$1120 at the end of the year. But if interest is compounded at the rate of 1 percent monthly (i.e. 1/12 of the annual rate), you get slightly more interest because it is compounded. Write a program which uses a for loop to compute the balance after a year of compounding interest in this way. The answer should be \$1126.83. Evaluate the formula for this result separately as a check.
money = 1000;
for month = 1:12
money = money*1.01;
end
disp(money)

check: 1000*1.0112

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.