C E numbers, commonly referred to as CE marking, signify that a product complies with the European Union's safety, health, and environmental protection standards. The CE marking is required for various products sold within the European Economic Area (EEA) and indicates that the manufacturer has assessed the product and ensured it meets all relevant EU directives. It facilitates free movement of products within the EU market while ensuring consumer safety.
C c d e e e e e d c f c c d d d d d c b ec a c c c c c a a c c c c c a a e e e e e e e e e e e e ....
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
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}]
The core of it would be something like this: int a,b,c,d,e,s; ... s=a+b+c+d+e;
#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.
D D# E C E C E C C D D# E C D E BD C D D# E C E C E C AG F# A C E D C A D D D# E C E C E C C D D# E C D E B D C C D C E C D E C D C E C D E C D C E C D E E B D C #=sharp
1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,10,11,12,13,14
1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,10,11,12,13,14