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c,c,f,f,f,f,c,f,f,f,f,f,c,f,f,e,d,c,c,c,e,e,e,e,e,e,e,e,e,e,e,d,c,c,d,e,f,c,c,f,f,c,c,c,f,f,f,c,c,c,f,f,e,d,c,c,c,e,e,e,e,e,e,e,d,c,c,d,e,f
C C C D E E E D C D E C E E F G G F E F G E C C D E E D C D E C G G C C C D E E E D C D E C
C(^), e, c b a g e, c c e, e, e d d c, e g, c c e, e, c d, d, c. E, g g e e g, f f f d b, g, c c e, e, c d, d, c. C(^) g c b g b a d a g. C c e, e, c d, d, c. (etc.)
C, E, G, E, C, E, G, E, A, C, E, C, A, C, E, C, F, A, C, A, F, A, C, A, F, F, F, F, F, F, F, F, F, F. C, E, G, E, C, E, G, E, A, C, E, C, A, C, E, C, F, A, C, A, F, A, C, A----, A. :) Hope this helps :)
Mean = sum of all numbers divided by number of numbers you summed. Call numbers a, b, c, d, e, f (a+b+c+d+e+f)/6 = mean
well, i think if you use this you can find out. A = 1-9 ,B = 0-9 , C = 0-9 , D = 0-9 , E = 0-9 for 2digit numbers = A A for 3 digit numbers = A B A for 4 digit numbers = A B B A and so on till you get to for 8 digit numbers = A B C D D C B A for 9 digit numbers = A B C D E D C B A and last for 10 digit number = A B C D E E D C B A this should work...
The sum of four numbers equals 40: a+b+c+d=40 Those same four numbers, along with another number (e), give the set a mean of 12: (a+b+c+d+e)/5=12 Use these two facts to determine e as follows: (a+b+c+d+e)/5=12 a+b+c+d+e=60 (a+b+c+d)+e=60 40+e=60 e=20
The core of it would be something like this: int a,b,c,d,e,s; ... s=a+b+c+d+e;
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
#include<stdio.h> #include<conio.h> void main() { int a,b,c,d,e; clrscr(); if(a>b) { if(a>c) e=a; else e=c; } else { if(b>c) e=b; else e=c; } if(e>d) printf("%d is the greatset",e); else printf("%d is the greatest",d); getch(); }
It is e: 29
jj
Let x = abc be a 3-digit number. The expanded for of x is 100*a + 10*b + c.And let y = de = 10*d + e be a 2-digit number. Then x*y = 1000*a*d + 100*a*e + 100*b*d + 10*b*e + 10*c*d + c*e = 1000*a*d + 100*(a*e + b*d) + 10*(b*e + c*d) + c*e.
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1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,10,11,12,13,14