Give the output of the following program (written in C syntax) using the four parameter-passing methods(Pass by Value, Pass by Reference, Pass by Value-Result, Pass by Name). Code: int i;int a[3]; void swap(int x, int y){    x = x + y;    y = x - y;    x = x - y;} main(){    i = 1;    a[0] = 2;    a[1] = 1;    a[2] = 0;    swap(i,a[i]);    printf("%d %d %d %d\n", i, a[0], a[1], a[2]);    swap(a[i],a[i]);    printf("%d %d %d\n", a[0], a[1], a[2]);    return 0;}

(Statistical) In many statistical analysis programs, data values considerably outside the range of the majority of values are simply dropped from consideration. Using this information, write a C++ program that accepts up to 10 floating-point values from a user and determines and displays the average and standard deviation of the input values. All values more than four standard deviations away from the computed average are to be displayed and dropped from any further calculation, and a new average and standard deviation should be computed and displayed.

When you borrow money to buy a house, a car, or for some other purpose, you repay the loan by making periodic payments over a certain period of time. Of course, the lending company will charge interest on the loan. Every periodic payment consists of the interest on the loan and the payment toward the principal amount. To be specific, suppose that you borrow $1,000 at an interest rate of 7.2% per year and the payments are monthly. Suppose that your monthly payment is $25. Now, the interest is 7.2% per year and the payments are monthly, so the interest rate per month is 7.2/12 = 0.6%. The first months interest on $1,000 is 1000 0.006 = 6. Because the payment is $25 and the interest for the first month is $6, the payment toward the principal amount is 25 6 = 19. This means after making the first payment, the loan amount is 1,000 19 = 981. For the second payment, the interest is calculated on $981. So the interest for the second month is 981 0.006 = 5.886, that is, approximately $5.89. This implies that the payment toward the principal is 25 5.89 = 19.11 and the remaining balance after the second payment is 981 19.11 = 961.89. This process is repeated until the loan is paid. Write a program that accepts as input the loan amount, the interest rate per year, and the monthly payment. (Enter the interest rate as a percentage. For example, if the interest rate is 7.2% per year, then enter 7.2.) The program then outputs the number of months it would take to repay the loan. (Note that if the monthly payment is less than the first months interest, then after each payment, the loan amount will increase. In this case, the program must warn the borrower that the monthly payment is too low, and with this monthly payment, the loan amount could not be repaid.)

(Practice) Write array declarations, including initializers, for the following: a. A list of 10 integer voltages: 89, 75, 82, 93, 78, 95, 81, 88, 77, and 82 b. A list of five double-precision slopes: 11.62, 13.98, 18.45, 12.68, and 14.76 c. A list of 100 double-precision distances; the first six distances are 6.29, 6.95, 7.25, 7.35, 7.40, and 7.42 d. A list of 64 double-precision temperatures; the first 10 temperatures are 78.2, 69.6, 68.5, 83.9, 55.4, 67.0, 49.8, 58.3, 62.5, and 71.6 e. A list of 15 character codes; the first seven codes are f, j, m, q, t, w, and z

(Practice) Write C++ statements for the following: a.b=sinxcosxb.b=sin2xcos2xc.area=( cbsina)/2d.c= a 2 + b 2 e.p= | mn|f.sum= a( r n 1) r1

(Practice) Write array declarations for the following: a. A list of 100 double-precision voltages b. A list of 50 double-precision temperatures c. A list of 30 characters, each representing a code d. A list of 100 integer years e. A list of 32 double-precision velocities f. A list of 1000 double-precision distances g. A list of 6 integer code numbers