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Ufumbuzi - Tiger algebra Suluhishaji

25867.59

25867.59

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (19220, 2, 1, 15)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 220$ , $r = 2 %$ , $n = 1$ , $t = 15$ .

$c i (19220, 2, 1, 15)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 220$ , $r = 2 %$ , $n = 1$ , $t = 15$ .

$P = 19220, r = 2 %, n = 1, t = 15$

$\frac{2}{100} = 0.02$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 220$ , $r = 2 %$ , $n = 1$ , $t = 15$ .

$A = P {(1 + r / n)}^{(n t)}$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 220$ , $r = 2 %$ , $n = 1$ , $t = 15$ .

$N e x t c o m p u t e r / n, n t, a n d {(1 + r / n)}^{(n t)} .$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 220$ , $r = 2 %$ , $n = 1$ , $t = 15$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.02$

$n t = 15$

${(1 + r / n)}^{(n t)}$

$r / n = 0.02, n t = 15, {(1 + r / n)}^{(n t)} = 1.3458683383$ .

${(1 + r / n)}^{(n t)}$

${(1 + 0.02)}^{15} = 1.3458683383$

$r / n = 0.02, n t = 15, {(1 + r / n)}^{(n t)} = 1.3458683383$ .

$1.3458683383$

$r / n = 0.02, n t = 15, {(1 + r / n)}^{(n t)} = 1.3458683383$ .

3. Hesabu kiasi cha mwisho

$A = 25867.59$

$25867.59$

$19, 220 \cdot 1.3458683383 = 25867.59$ .

$25867.59$

$19, 220 \cdot 1.3458683383 = 25867.59$ .

$25867.59$

$19, 220 \cdot 1.3458683383 = 25867.59$ .

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Tiger algebra Suluhishaji

Ufumbuzi - Tiger algebra Suluhishaji

Maelezo kwa hatua

1. Weka fomula ya riba changamano

2. Hesabu kipengele cha ukuaji

3. Hesabu kiasi cha mwisho

Kwa nini kujifunza hii

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