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problem)Atmospheric carbon dioxide is thought to be responsible for global warming.
The table below contains data about the amount of Atmospheric CO2 (in ppm) over the last fifty years.

Year    CO2
1959    315.98
1960    316.91
1965    320.04
1968    323.04
1971    326.32
1975    331.08
1980    338.7
1982    341.4
1984    344.4
1986    347.2
1988    351.5
1990    354.2
1992    356.4
1994    358.9
1996    362.6
1998    366.6
2000    369.4
2002    372.9
2007    383.71
2008    385.54
2009    387.35

(a) Construct a scatter plot for this data. (Either by hand or using Excel)
(b) Comment on the shape.
(c) Find a line of best fit.  If doing this by hand, you will need to find the line of best fit by eye and then find the equation.  If using excel, use a formula(s) to find the line of best fit.
(d) Using the equation, what was/is the atmospheric CO2 in 2005, 1920, 2020?  Comment on the accuracy of your answers.
(e) In your conclusion, comment on the accuracy of modeling this as a straight line given your graph.  Consider the accuracy to determine past and present values for CO2.
Data sourced from http://co2now.org/index.php?option=com_content&task=view&id=22&Itemid=1
problem)Forensic Science – Bone Lengths

Forensic Scientists can estimate the height of deceased people from the length of a few key bones.
Heights can be determined from the length of the humerus, (the bone that extends from the elbow to the shoulder) using the equations:
For males: H= 3.08h + 70.45, For females H=3.36h + 57.97

Where H is the height of the deceased and h is the length of the humerus in cm.
Heights can also be estimated from the length of the tibia, (the larger bone that extends from just below the kneecap to the ankle) using the equations:
For males: H=2.42t + 81.93, for females H=2.9t + 61.53

Where H is the height of the deceased and t is the length of the tibia in cm.
Heights can also be estimated from the length of the ulna, (the larger bone that extends from the elbow to the hand) using the equations:

For males: H=3.7u + 70.45, for females H=4.27u + 57.76

Where H is the height of the deceased and u is the length of the ulna in cm.
What is the height of the person and are they male or female?

Complete the following:
(a) A tibia of length 32 cm is found in a burial site, find the person’s height? (the sex is unknown)
(b)A collection of bones is found where the humerus = 34cm, tibia = 38.5cm and ulna = 28.3cm.  Was this person a male or female

Based on your height, what will the length of your humerus, tibia and ulna be?

problem) Tyre Life

Data about tyre wear was collected. The number of kilometres from Brand A tyres is given below:

The number of kilometres from Brand A tyres is:

35000     38000     38000     42000     45000
45000     49000     50000     50000     52000
53000     58000     62000     62000     63000
66000     67000     70000     71000     74000
76000     81000     81000     86000     94000

The number of kilometres from Brand B tyres is:

46000     49000     55000     56000     58000
64000     64000     64000     67000     68000
68000     69000     69000     72000     72000
74000     76000     78000     79000     79000
82000     86000     87000     94000     97000

Which tyre gives the longest life?
(a) Draw a side by side box and whisker plot to determine the answer. (Hint: arrange the data in a back to back stem and leaf plot first)
problem 3 Male and Female Athletes
The table below contains times for males and females times for 100m running races at Olympic Games.
Year    Years since 1928    Men’s 100m time (sec)    Women’s 100m time (sec)
1928    0                                   10.8                              12.2
1932    4                                   10.3                              11.9
1936    8                                   10.3                              11.5
1948    20                                  10.3                              11.9
1952    24                                  10.4                              11.5
1960    32                                  10.2                              11.0
1968    40                                  9.95                              11.08
1976    48                                  10.06                             1.08
1980    52                                  10.25                            11.06
1992    64                                  9.96                              10.82
1996    68                                  9.84                              10.94
2000    72                                  9.87                              10.75
2004    76                                  9.85                              10.93
2008    80                                  9.69                              10.78
Construct a graph for this data. (Either by hand or using Excel)
Determine the equation for the line of best fit by both calculation and using excel. Compare the two answers.
Will female athletes ever catch up to male athletes? When might this happen?
problem 4   Household Water Use
1. This is a roof plan (from above) of a house.

2018_roof plan.jpg

(a)During a recent rainy day, 72 mm of rain was recorded.  find out how many litres of water will be collected from the roof and stored in the water tank. Remember  1cm3 = 1mL or 1m3 = 1kL(1000L)
The volume V of water in the tank, initially full, remaining after (d) days is given by the equation:
         V=10 000-235 d

(b)     How much is left from a full tank after 30 days of no rain?
(c)     From the equation, determine how many litres of water the tank holds when full.
(d)     From the equation, determine how many litres of water the householders use per day
(e)     When will the volume of the tank be reduced to 2500 litres?

problem 5 Break-even analysis
A tourist coach company offers a daily (week days) tour from Bryon Bay to Lismore and return. People taking the tour are supplied lunch
The company puts a weekly advertisement in the local paper costing $120.
Pc = $120.00

Each passenger is charged $45 (including lunch) for the return journey.
Ep=$45.00

The fuel cost and other running costs total $45 per trip.
F=$45.00

The driver’s wage is $120 per trip.
Dw=$120.00

Lunch cost is $14 per person
L=$14.00

Outlay = Pc=$120c a week, F=$45, Dw=$120= $299.00 Each trip costs= $285
Variable =, Lunch=$14
Income = Per passage Ep=$45 ($30)
P=$120+$120+$45=$285

If clients pay $45 and this cost is called $31 then this removes the variable in the equation of the $14 for lunch.
10 clients per trip to make a profit
The company would like to make a profit on a weekly basis even though some individual trips might operate at a loss.

(a) From the information given, plot the graphs of revenue and cost and find the number of passengers required per week to break even.
(b) Ideally the company would like to make $1000 profit per week, find out the number of passengers required to make this happen. 
(c) What is the maximum profit possible if the bus can hold a maximum of 20 passengers per trip?

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  • Category:- Math
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