0
You can buy a especific kit for your Clio. Go to the k&n site and search for your car. There is also information in pictures, how to fit the kit in your engine.
I have tried it on my 1997 1.2 Clio.
Wiki User
∙ 15y ago
Add your answer:
Earn +20 pts
When did Jagathy N. K. Achary die?
Jagathy N. K. Achary died in 1997.
What is the firing order for a Renault Clio 1994?
firing order is 1243 N reg 1.4 Clio
How do you solve a permutation in math?
If there are n objects to fit r places (e.g. 9 people in 7 chairs, 4 tumblers in a lock) then the number of permutations is nCk, stated as n-choose-k. This number can be calculated by the formula n!/(n - k)!. If k is equal to n, then (n - k)! = 0! = 1, and the number of permutations is simply n!. If the direction of the permutation is irrelevant (e.g. ABCD is the same as DCBA) then divide by two to cancel out the double-counting.
Will any k and n cold air intake fit a ford ranger?
no they have specific models for specific years and engines. go to k and n website or your local parts dealer and have them look up which one you will need
C program for displaying perfect numbers using nested loop?
for (n=1; n<1000; ++n) { for (sum=0, k=1; k<=n/2; ++k) if (n%k==0) sum += k; if (sum==n) printf ("%d\n", n); }
Does a PH16 Fram oil filter fit a 2011 Dodge Dakota 4.7L?
Yes, but i would use a mopar or k&n.
How do you write it in numbers k is a function of n and 7 times n equals k?
k = f(n) = 7n
Would a 1997 Toyota supra engine and trans fit n a 1990 Mazda miata?
Unfortunately, a 1997 Toyota Supra engine and trans would not fit into a 1990 Mazda Miata. The Supra engine and transmission are unable to be mounted correctly within a Miata to make them fit.
C program for optimal merge pattern?
#include<iostream.h> #include<conio.h> void main() { clrscr(); int i,k,a[10],c[10],n,l; cout<<"Enter the no. of elements\t"; cin>>n; cout<<"\nEnter the sorted elments for optimal merge pattern"; for(i=0;i<n;i++) { cout<<"\t"; cin>>a[i]; } i=0;k=0; c[k]=a[i]+a[i+1]; i=2; while(i<n) { k++; if((c[k-1]+a[i])<=(a[i]+a[i+1])) { c[k]=c[k-1]+a[i]; } else { c[k]=a[i]+a[i+1]; i=i+2; while(i<n) { k++; if((c[k-1]+a[i])<=(c[k-2]+a[i])) { c[k]=c[k-1]+a[i]; } else { c[k]=c[k-2]+a[i]; }i++; } }i++; } k++; c[k]=c[k-1]+c[k-2]; cout<<"\n\nThe optimal sum are as follows......\n\n"; for(k=0;k<n-1;k++) { cout<<c[k]<<"\t"; } l=0; for(k=0;k<n-1;k++) { l=l+c[k]; } cout<<"\n\n The external path length is ......"<<l; getch(); }
Sum of 1 plus 2 plus 3 plus 4 plus .. plus n?
n(n+1)/2 You can see this from the following: Let x=1+2+3+...+n This is the same as x=n+(n-1)+...+1 x=1+2+3+...+n x=n+(n-1)+...+1 If you add the corresponding terms on the right-hand side of the two equations together, they each equal n+1 (e.g., 1+n=n+1, 2+n-1=n+1, ..., n+1=n+1). There are n such terms. So adding the each of the left-hand sides and right-hand sides of the two equations, we get: x+x=(n+1)+(n+1)+...+(n+1) [with n (n+1) terms on the right-hand side 2x=n*(n+1) x=n*(n+1)/2 A more formal proof by induction is also possible: (1) The formula works for n=1 because 1=1*2/2. (2) Assume that it works for an integer k. (3) Now show that given the assumption that it works for k, it must also work for k+1. By assmuption, 1+2+3+...+k=k(k+1)/2. Adding k+1 to each side, we get: 1+2+3+...+k+(k=1)=k(k+1)/2+(k+1)=k(k+1)/2+2(k+1)/2=(k(k+1)+2(k+1))/2=((k+2)(k+1))/2=(((k+1)+1)(k+1))/2=((k+1)((k+1)+1)/2
Write a program in c language to print a diamond?
// // THIS IS A MACH SIMPLER SOLUTION: // void Diamond(int n) { for (int i=0;i<=2*n;i++,printf("\n")) for (int j=0;j<=2*n;j++) (abs(i-n)+abs(j-n)<=n ? printf("*") : printf(" ")); } //============================================= #include<stdio.h> main() { int i,j,k,n,a,b,c,x; printf("enter the # of rows of graphical output"); scanf("%d",&n); /* UPPER HALF OF KITE */ for(i=1;i<=n;i++) { printf("\t"); for (k=1;k<=(n-i);k++) { printf(" "); } for(j=0;j<i;j++) { printf("*"); printf(" "); } for(k=1;k<=(n-i-1);k++) { printf(" "); } printf("\n"); } /* LOWER PART OF KITE */ for(i=(n-1);i>0;i--) { printf("\t"); for (k=(n-i);k>0;k--) { printf(" "); } for(j=i;j>0;j--) { printf("*"); printf(" "); } for(k=(n-i-1);k>0;k--) { printf(" "); } printf("\n"); } getch(); }
If a fair coin is tossed eight times what is the probability of getting exactly 6 heads?
You need a formula for this. If the probability (in one toss) of getting head is "p", then the probability of getting exactly k heads out of n tosses is: (n,k) p^k (1-p)^(n-k) where (n,k) denotes the number of combinations of k elements among n. You should also know that (n,k) = n! / (( n-k)! k! ) so here, with n=8, k=6, and p=.5 you have (n,k) = 8*7 / 2 = 28 and your probability is : 28 * 1/2^6 * 1/2^2 = 28 / 256 = 7 / 64
