(a) Find the constant change in population per year of a city from 2008 to 2012, given an increase from 23,400 to 27,800. (b) Determine the linear cost function $C(x)$ for producing $x$ items with a fixed cost of K1,250 and a production cost of K37.50 per item, then find the cost for producing 100 items. (c) Estimate the chi-square value for $n = 23$ and $p = 99$ based on a provided table. (d) Calculate the number of possible car registration plates in PNG consisting of 3 letters followed by 3 digits, with no repetition of characters allowed.

Applied MathematicsLinear FunctionsStatisticsCombinatoricsPopulation Growth
2025/4/6

1. Problem Description

(a) Find the constant change in population per year of a city from 2008 to 2012, given an increase from 23,400 to 27,
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0

0. (b) Determine the linear cost function $C(x)$ for producing $x$ items with a fixed cost of K1,250 and a production cost of K37.50 per item, then find the cost for producing 100 items.

(c) Estimate the chi-square value for n=23n = 23 and p=99p = 99 based on a provided table.
(d) Calculate the number of possible car registration plates in PNG consisting of 3 letters followed by 3 digits, with no repetition of characters allowed.

2. Solution Steps

(a)
The population increased from 23,400 to 27,800 between 2008 and
2
0
1

2. The time period is 2012 - 2008 = 4 years.

The total increase in population is 27,800 - 23,400 = 4,
4
0

0. The average increase per year is 4,400 / 4 = 1,

1
0
0.
(b)
The linear cost function C(x)C(x) can be written as C(x)=fixed cost+(cost per item)×xC(x) = \text{fixed cost} + (\text{cost per item}) \times x.
Given the fixed cost of K1,250 and the production cost of K37.50 per item, the cost function is C(x)=1250+37.50xC(x) = 1250 + 37.50x.
To find the monthly cost for producing 100 items, substitute x=100x = 100 into the cost function:
C(100)=1250+37.50(100)=1250+3750=5000C(100) = 1250 + 37.50(100) = 1250 + 3750 = 5000.
(c)
The chi-square table provides values for n=20,22,24n = 20, 22, 24 and p=95,99,99.9p = 95, 99, 99.9. We need to estimate the value for n=23n = 23 and p=99p = 99.
First, find the values for p=99p = 99 when n=22n = 22 and n=24n = 24 which are 40.2940.29 and 42.9842.98.
Then, find the mean of n=22n=22 and n=24n=24 for p=99p=99, which corresponds to the estimated chi-square value for n=23n=23, p=99p=99.
Average 40.2940.29 and 42.9842.98, (40.29+42.98)/2=83.27/2=41.635(40.29+42.98)/2 = 83.27/2 = 41.635.
(d)
In PNG, car registration plates have 3 letters followed by 3 digits. Repetitions are not allowed.
There are 26 letters in the alphabet.
The first letter has 26 options.
The second letter has 25 options (since repetitions are not allowed).
The third letter has 24 options.
There are 10 digits (0 to 9).
The first digit has 10 options.
The second digit has 9 options.
The third digit has 8 options.
The total number of possible car registration plate numbers is 26×25×24×10×9×8=11,232,00026 \times 25 \times 24 \times 10 \times 9 \times 8 = 11,232,000.

3. Final Answer

(a) The population increased by 1,100 per year.
(b) The linear cost function is C(x)=1250+37.50xC(x) = 1250 + 37.50x. The monthly cost for producing 100 items is K
5
0
0

0. (c) The estimated chi-square value for $n=23$ and $p=99$ is 41.

6
3

5. (d) The number of possible car registration plate numbers is 11,232,

0
0
0.

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