Transcript File - Ms Burton`s Weebly

Revision
Question 1
A person who initially weighs 120.0kg begins a diet.
The first month they loose 18.0kg, the next month
12.0kg and the following month 8.0kg and they
continue to lose weight by decreasing by the same
ratio each successive month
1. Give a formula that represents the amount of
weight lost in a particular month ‘n’ and find out how
much weight the person would be expected to loose in
month 10
2. How much would the person weigh after 10 months
on the diet
3. If the person continued on the diet indefinitely
what would be the maximum amount of weigh they
could loose and their final weight?
Give a formula that represents the amount of
weight lost in a particular month ‘n’ t = a  rn-1
n
tn = 18  (2/3)n-1
And, find out how much weight the person would be
expected to loose in month 10t10 = 18  (2/3)9
= 0.468kg
2. How much would the person weigh after 10 months
on the diet S10 = 18 x (1- (2/3)10) = 53.1kg
1.
(1 – 2/3)
Weight of person after 10 months is 120 – 53.1 = 66.9kg
3. If the person continued on the diet indefinitely
what would be the maximum amount of weight
they could loose and their final weight?
S∞ = 18
(1 – 2/3)
= 66. The maximum amount of weight they could loose is 66kg
Their final weight would be (120kg – 66) = 54kg
Question 2
Manu increases the amount of laps she swims
daily by the same number. After 7 days she has
swum 154 laps altogether and after 21 days has
swum 1050 laps.
1. How many laps did she swim on the first day?
Manu swum 10 laps on her first day
2. When will she reach her target of a total of
2000 laps swum
Sn = 2000, n = 29.685 or -33.685
(ignore the negative solution)
On the 30th day Manu will achieve her target of a total of
2000 laps swum.
Note: after 29 days Manu has only swum 1914 laps and so it
is on the 30th day she passes the 2000 lap mark
Question 2 – part 1
1. How many laps did she swim on the first day?
n
Using the formula Sn = 2 2a  (n  1)d
7é
2a + (7 - 1)dù = 154
S7 = 154, so
û
2ë
7a + 21d = 154
S21 = 1050, so
Equation 1
21 é
2a + (21 - 1)dù = 1050
û
2ë
21a + 210d = 1050 Equation 2
We now have 2 equations with 2 unknown variables a and d
We can solve these on the calculator using Simultaneous equations
EQUA – F1(Simult) – F1(number of unknowns = 2 which are a and d)
Note: your equations must be in the correct order
7a + 21d = 154
21a + 210d = 1050
Equation 1
Equation 2
We now have 2 equations with 2 unknown variables a and d
We can solve these on the calculator using Simultaneous equations.
EQUA – F1(Simult) – F1(number of unknowns = 2 which are a and d)
Note: your equations must be in the correct and same order
anX + bnY = Cn
Enter in the numbers of equation 1 and 2
7 EXE 21EXE 154 EXE 21 EXE 210 EXE 1050 EXE
Then press F1(Solve)
You will get x[10]
y[4 ]
Relating to our our equation a=x and d = y
Therefore a (first term) = 10 and d (common difference) = 4
Question 2 – part 2
Manu increases the amount of laps she swims daily by
the same number. After 7 days she has swum 154 laps
altogether and after 21 days has swum 1050 laps.
2. When will she reach her target of a total of 2000
laps swum
Using the formula Sn =
2000 =
2000 =
2000 =
n
2a  (n  1)d

2
Now know that a = 10 and d = 4
né
2 ´ 10 + (n - 1) ´ 4ù
ë
û
2
né
20 + 4n - 4ù
ë
û
2
né
16 + 4nù
ë
û
2
2000 = 8n + 2n2 Need to have this written in the form ax2 + bx + c = 0
to solve on the calculator
2n2 + 8n - 2000 = 0
Question 2 – part 2
2000 = 8n + 2n2
2n2 + 8n - 2000 = 0
Need to have this written in the formax2 + bx + c = 0
to solve on the calculator
We can solve this on the calculator using Polynomial.
EQUA – F2(Poly) – F1(Degree ? The degree is 2 – the highest power in the
polynomial ie. n2 has a degree of 2)
Note: your equations must be in the correct and same order as
ax2 + bx + c = 0
Enter in the numbers of equation
2 EXE 8 EXE -2000 EXE
Then press F1(Solve)
You will get x1 [29.685]
x2[-33.68 ]
Therefore n = 29.685 or -33.685 (ignore the negative solution)
Note 1: after 29 days Manu has only swum 1914 laps and so it is on the 30th day
she passes the 2000 lap mark
On the 30th day Manu will achieve her target of a total of 2000 laps swum.
Question 3
A computer artificial intelligence system is designed to learn from
its own mistakes. When it was trialled on understanding some
written English, it made 850 errors.
r = 0.95
Every trial after this it makes only 95% of the errors of the
previous trial.
After how many trials will the number of errors drop to 100 or
less?
a = 850
tn = a  rn-1
r = 0.95
tn = 850  (0.95)n-1
100 > 850  (0.95)n-1
0.117647 > (0.95)n-1
n = 43
using solver (F3)
Remember to put brackets around the power!!
=850 x 0.95⌃(n-1) EXE F6(solv)
= 42.722
100