IDNStudy.com, kung saan ang kuryusidad ay nagtatagpo ng kalinawan. Magtanong at makatanggap ng eksaktong sagot mula sa aming mga bihasang miyembro ng komunidad.

Send answer senior high students asap

I'll send load worth P100.00 if you can give me correct answer promise.




Send Answer Senior High Students AsapIll Send Load Worth P10000 If You Can Give Me Correct Answer Promise class=

Sagot :

1. A = P(1+rt)

= 50000(1+5%(3))

= 50000(1+0.05(3))

= 50000(1+(0.15))

= 50000(1.15)

= P75000

2. A = P(1+rt)

= 20000(1+2%(2))

= 20000(1+0.02(2))

= 20000(1+(0.04))

= 20000(1.04)

= P20800

3. A = P(1+r/n)^(nt)

= 400000(1+2%/4)^(4(12))

= 400000(1+0.02/4)^(48)

= 400000(1.005)^(48)

= P508195.66

4. A = P(1+r/n)^(nt)

50000 = 10000(1+6%/2)^(nt)

50000 = 10000(1+0.06/2)(2t)

50000 = 10000(1+1.03)^(2t)

50000 = 10000(2.03)^(2t)

50000 = (4.120)^(t)

4.12^t = 50000

t = 7.64 years

5. Let's say P = 10000

1st given

• A = P(1+rt)

= 10000(1+0.10(1.5)

= 10000(1.15)

= 11500

= P11536.89

2nd given

• A = P(1+r/n)^(nt)

= 10000(1+0.095/12)^(12(1.5)

= P11525.06

Kat should choose the 9.5% compounded monthly for 1½years

6. A = P(1+r/n)^(nt)

= 10000(1+0.02/3)^(3(5))

= P11048.04

7. F = P([1+i]^n-1)/i

= 18000([1+0.1]^12-1)/0.1

= 384917.10

= 384917.10+180000

= P564917.10

8. F = P([1+i]^n-1)/i

1000000 = P([1+0.04]^15(2)-1)/0.04

P = P17830.09

9. F = P([1+i]^n-1)/i

1000000 = P([1+0.05]^1(15) -1/0.05

P = P46342.28

10. P = P[1-(1+i)^-n/i]

= 5500[1-(1+0.12)^-(12(2))/0.12]

= P42813.73