f3 mth questions!!!急~聽日要交!

Q1:
4(1-50%)^19200/1600
0.000976563kg
0.976g
1g

2010-11-06 20:44:27 補充：
$8,340,000(k%)=$8,737,000
k%=1.0476
k=105(cor. to nearest integer)

it means the growth factor of k is 105%
______________
2002:
x(104.76%)=$8340000
x=7960000(cor. to 3 sig. figs.)

2006:
8,737,000(104.76%)^2=$9 590 000(cor. to 3 sig. figs.)

2010-11-06 20:58:36 補充：
Let x be the saving of May:

June:
0.9x

July:
0.81x

x+0.9x+0.81x=40650
x=15000

2010-11-06 20:59:18 補充：
Q3好有誤導成分.......

2010-11-06 21:29:06 補充：
Q1:
4(1-50%)^19200/1600
0.000976563kg
0.976g
1g

2010-11-06 21:29:34 補充：
$8,340,000(k%)=$8,737,000
k%=1.0476
k=105(cor. to nearest integer)

it means the growth factor of k is 105%
______________
2002:
x(104.76%)=$8340000
x=7960000(cor. to 3 sig. figs.)

2006:
8,737,000(104.76%)^2=$9 590 000(cor. to 3 sig. figs.)

2010-11-06 21:29:51 補充：
Let x be the saving of May:

June:
0.9x

July:
0.81x

x+0.9x+0.81x=40650
x=15000

7. The weight of a radioactive chemical decreases by 50% every 1,600 years. If its original weight is 4 kg ,what will its weight be after 19,200 years?(give the ans correct to the nearest g.)ANS : 1g

19200 years = 12 * 1600 years

decreases by 50% = 1/2

4kg = 4*1000 g

Its weight willbe after 19,200 years :

4*1000 * ( 1/2 ) ^12

4000 * 1/4096

0.9765625

~ 1g ( give the ans correct to the nearest g.)

18.The following table shows the values of a flat in 2003 and 2004.YearValue2003$8,340,0002004$8,737,000 Suppose the value of this flat increases at a constant rate year, and the growth factor of k.(a) find the value of k.
k = 1 + ( 8737000 - 8340000 ) / 8340000

k = 1.047601918465227817745803357314149

~ 1.05

ANS : 1.05

(b) Find the value of the flat in the following years :(i) 2002

= 8340000 / 1.047601918465227817745803357314149

= 7961039.2583266567471672198695216...

~ 7960000

(ii) 2006
= 8,737,000 * 1.047601918465227817745803357314149^2

= 9588593.4641207896991759110697027...

~ 9590000

= (give the ans correct to 3 sig.fig.) (i) ans : $7 960 000 (ii) ans : $9 590 000