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What do you mean by P1 rating in bonds and debentures?

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Answered 2008-05-28 09:51:37

P1" is the highest short-term rating category for Moody's Investor Service. P1 rating are considered to be of high credit quality

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Asterisks indicate a section of the instructions that will be repeated. For example, with the instruction below:K2, *P1, K1* repeat 2 times, P1, K2You will actually be doing this:K2, P1, K1, P1, K1, P1, K1, P1, K2 (the bolded stitches are the ones from inside the asterisks)


P1 refers to the parental generation. Successive generations (Filial generations) are labeled F1, F2, etc.


Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2


reverse (char *p) { char c, *p1, *p2; if (strlen(p) <= 1) return; for (p1=p, p2=p+strlen(p)-1; p1<p2; c=*p1, *p1=*p2, *p2=c, p1++, p2--); }


The 'arrow' means :Reactants 'are reacting to form' ProductsR1 + R2 --> P1 + P2R1 and R2 are reacting to form P1 and P2


C program for Shortest Job First Preemptive algorithm.#includestruct proc{int pid;int at,bt,wt,tat,rbt;int flag,flag1;};struct proc p1[10];int i,j,k,n,no,m;float atat=0.0,awt=0.0;int tbt=0;int minimum1();int main(){int minv,locv,mins,locs;printf("\nenter the number of processes:");scanf("%d",&n);printf("\nenter the proc information:");printf("\npid at bt");for(i=0;i{p1[i].wt=0;p1[i].tat=0;p1[i].flag=0;p1[i].flag1=0;scanf("%d%d%d",&p1[i].pid,&p1[i].at,&p1[i].bt);tbt+=p1[i].bt;p1[i].rbt=p1[i].bt;}printf("\nthe proc information:");printf("\npid at bt");for(i=0;i{printf("\n%d %d %d",p1[i].pid,p1[i].at,p1[i].bt);}minv=p1[0].at;locv=0;for(i=1;i{if(p1[i].at{locv=i; //tells min at process in locvminv=p1[i].at;}}for(i=0;i{if(p1[i].at==minv){p1[i].flag1=1; //processes having same minimum at}}mins=p1[0].bt;locs=0;for(i=0;i{if(p1[i].flag1==1&&p1[i].bt{mins=p1[i].bt; //gives process with minimum burst timelocs=i;}}printf("\ngantt chart:");for(i=minv;i{for(j=0;j{if(p1[j].rbt>0&&p1[j].at{p1[j].flag=1;}}no=minimum1();printf("%d p[%d]",i,p1[no].pid);p1[no].rbt=p1[no].rbt-1;for(k=0;k{if(p1[k].rbt>0&&p1[k].at{p1[k].wt++;}}}printf("%d",tbt+minv);for(i=0;i{awt+=p1[i].wt;}awt=awt/n;for(i=0;i{p1[i].tat=p1[i].wt+p1[i].bt;atat+=p1[i].tat;}atat=atat/n;printf("\n average wt=%f, average tat=%f",awt,atat);printf("\nthe proc information:");printf("\npid at bt wt tat");for(i=0;i{printf("\n%d %d %d %d %d",p1[i].pid,p1[i].at,p1[i].bt,p1[i].wt,p1[i].tat);}}int minimum1(){int loc,z;int mini;mini=99;loc=-1;for(z=0;z{if(p1[z].rbt>0&&p1[z].at{mini=p1[z].rbt;loc=z;}}return loc;}


char *strcpy(char *p1, const char *p2) char *pt = p1, while (*(p1++) = *(p2++)), return pt; ... or, at warning level 4 ... char *strcpy(char *p1, const char *p2) char *pt = p1, while ((*(p1++) = *(p2++)) != '\0'), return pt;


P1-e is an expression, not a formula.


In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation


If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%


I think it means PLACE MARKER in the row of stitches where instructed. If purling, it would read, P1,PM,P1, etc. (Purl 1, place marker, purl 1.)


P1 = V I1, Therefore, if P2=0.5*P1, thenI2=0.5*P1/V, or 0.5*I1and if P3=2*P1, thenI3=2*P1/V, or 2*I1In other words, current is proportional to power and inversely proportional to voltage.


P1 Material is the name given to Carbon Manganese Steel in the engineering industry



Consider two stacks p1 and p2 of same size.Enqueue:Push element into p1. If p1 is full, then pop all elements from p1 and push it into p2. Now push the new element into p1.Dequeue:Pop element from p2. If p2 is empty, the pop all elements from p1 and push it into p2. Now pop element from p2.


#include#include#define n 100void del_char(char str1[n],char str2[]){char str[n],*p1,*q,i,j;p1=str1;q=str;*q=*p1;i=0;while(*p1!='\0'){{if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;}else{q++;}}else{q++;}}else{p1++;q++;}*q=*p1;}i=0;}printf("%s\n",str);}main(){char str1[n],str2[]={"to "};clrscr();printf("please enter a text and includ 'to':\n");gets(str1);printf("remove 'to',remaining :\n");del_char(str1,str2);getch();return 0;}Output:please enter a text and includ 'the':Write a c program to simulate the calculator using functionremove 'the',remaining :Write a c program to simulate calculator using function


char *p1 = "Hello"; char *p2 = "World"; char *temp = malloc (sizeof(p1) + sizeof(p2)); i=0; while(*p1) { temp[i] = *p1++;i++; } while(*p2){ temp[i] = *p2++;i++; } temp[i] = '\0';


It's most likely PL as in plano, Which means there is no correction.


Assignment. Eg: void *p1, *p2; p2= p1;


There is no P1. It's either "P", or nothing... after Pit's F1 then F2...


The F1 generation or (first filial) indicates that it is the first offsring of the P1 generation.


If you think to barium phosphide the correct formula is Ba3P2.


Price elasticity demand formula end point formula epd= [q2-q1/q1]/[p2-p1/p1] midpoint formula epd= [q2-q1/(q2+q1)/2] / [p2-p1/(p2+p1)/2]


No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k.  Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.


B rhesus positive anti-P1 negative



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