(1)复利现值系数表 n 1% 2% 5% 6% 10% 12% 15% 1 0.9901 0.9804 0.9524 0.9434 0.9091 0.8929 0.8696 2 0.9803 0.9612 0.9070 0.8900 0.8264 0.7972 0.7561 3 0.9706 0.9423 0.8638 0.8396 0.7513 0.7118 0.6575 4 0.9610 0.9238 0.8227 0.7921 0.6830 0.6355 0.5718 5 0.9515 0.9057 0.7835 0.7473 0.6209 0.5674 0.4972 6 0.9420 0.8880 0.7462 0.7050 0.5645 0.5066 0.4323 10 0.9053 0.8203 0.6139 0.5584 0.3855 0.3220 0.2472(2)普通年金现值系数表n 1% 2% 5% 6% 10% 12% 15% 1 0.9901 0.9804 0.9524 0.9434 0.9091 0.8929 0.8696 2 1.9704 1.9416 1.8594 1.8334 1.7355 1.6901 1.6257 3 2.9410 2.8839 2.7232 2.6730 2.4869 2.4018 2.2832 4 3.9020 3.8077 3.5460 3.4651 3.1699 3.0373 2.8550 5 4.8534 4.7135 4.3295 4.2124 3.7908 3.6048 3.3522 6 5.7955 5.6014 5.0757 4.9173 4.3553 4.1114 3.7845 10 9.4713 8.9826 7.7217 7.3601 6.1446 5.6502 5.0188

(1)复利现值系数表 n 1% 2% 5% 6% 10% 12% 15% 1 0.9901 0.9804 0.9524 0.9434 0.9091 0.8929 0.8696 2 0.9803 0.9612 0.9070 0.8900 0.8264 0.7972 0.7561 3 0.9706 0.9423 0.8638 0.8396 0.7513 0.7118 0.6575 4 0.9610 0.9238 0.8227 0.7921 0.6830 0.6355 0.5718 5 0.9515 0.9057 0.7835 0.7473 0.6209 0.5674 0.4972 6 0.9420 0.8880 0.7462 0.7050 0.5645 0.5066 0.4323 10 0.9053 0.8203 0.6139 0.5584 0.3855 0.3220 0.2472

(2)普通年金现值系数表

n 1% 2% 5% 6% 10% 12% 15% 1 0.9901 0.9804 0.9524 0.9434 0.9091 0.8929 0.8696 2 1.9704 1.9416 1.8594 1.8334 1.7355 1.6901 1.6257 3 2.9410 2.8839 2.7232 2.6730 2.4869 2.4018 2.2832 4 3.9020 3.8077 3.5460 3.4651 3.1699 3.0373 2.8550 5 4.8534 4.7135 4.3295 4.2124 3.7908 3.6048 3.3522 6 5.7955 5.6014 5.0757 4.9173 4.3553 4.1114 3.7845 10 9.4713 8.9826 7.7217 7.3601 6.1446 5.6502 5.0188


相关考题:

有以下程序struct S{ int n; int a[20]; };void f(int *a,int n){ int i;for(i=0;in-1;i++) a[i]+=i;}main( ){ int i; struct S s={10,{2,3,1,6,8,7,5,4,10,9}};f(s.a,s.n);for(i=0;is.n;i++) printf("%d",s.a[i]);}程序运行后的输出结果是A)2,4,3,9,12,12,11,11,18,9,B)3,4,2,7,9,8,6,5,11,10,C)2,3,1,6,8,7,5,4,10,9,D)1,2,3,6,8,7,5,4,10,9,

以下选项中关于终值和现值的说法不正确的是()。A.1/(1+i)^n称为复利现值系数或l元的复利现值,用符号(PV,i,n)表示B.(1+i)^n称为复利终值系数或1元的复利终值,用(FV,i,n)表示。例(FV,6%,3)表示利率为6%,3期复利终值系数C.复利终值的计算公式为FV=PV×(1+i)^nD.单利是指每经过一个计息期,要将所生利息加入本金再计利息的方法

有以下程序#includestdio.h#define N 4void fun(int a[][N],int b[]){ int i;for(i=0;iN;i++)b[i]=a[i][i]-a[i][N-1-i];}main(){ int x[N][N]:{{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}},y[N],i;fun(x,y);for(i=0;iN;i++) prinff("%d,",y[i]);printf("\n");}程序运行后的输出结果是A.-12,-3,0,0,B.-3,-1,1,3,C.0,1,2,3,D.-3,-3,-3,-3,

试题28有以下程序#include stdio.h#define N 4void fun(int a[][N], int b[]){ int I;for(i=0; iN; i++) b[i]=a[i][i]-a[i][N-1-i];}main(){ int x[N][N]={{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}},y[N], i;fun(x,y);for(i=0;iN;i++) printf(“%d,”, y[i]); printf(“\n”);}程序运行后的输出结果是()A.-12,-3,0,0,B.-3,-1,1,3C.0,1,2,3,D..-3,-3,-3,-3

在定义int n[5][6];后,若认为n[0][0]为第1个元素,则第10个元素是________。A.n[2][5]B.n[2][4]C.n[1][3]D.n[1][4]

(1+i)-n为复利现值系数,或1元的复利现值,用(F/P,i,n)表示。

逻辑函数F=AB+BC+CD写成最小项之和,其结果应为∑m(________);写成最大项之积,其结果应为∏M(________)。A.(3,6,7,10,12,13,14,15);(0,1,2,4,5,8,9,11)B.(3,6,7,11,12,13,14,15);(0,1,2,4,5,8,9,10)C.(3,6,7,11,14,15);(0,1,2,4,5,8,9,10)D.(3,6,7,11,12,13,14,15);(0,1,2,8,9,10)

在利率相同的情况下,第8年年末1元的复利现值系数大于第6年年末1元的复利现值系。 ()

以下程序段的运行结果是()。 int a[]={1,2,3,4,5,6,7},*p=a; int n,sum=0; for(n=1;n<6;n++) sum+=p[n++]; printf("%d",sum);A.12B.15C.16D.27