MATH6000: Week 6

I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

Assessment 2 Results

Pending….

Week 6

DCOM

We finished talking about statistics and then started the third chapter on algebra.

WEB/DNET

We finished talking about statistics and then started the third chapter on algebra.

Week 7

We will start talking about equations.

Tutorials

At the moment we are looking at these exercises.

DCOM1 Group A: Wednesday 09:00 in C212 (with Leonard O’Driscoll) and Thursday 09:00 in B185 (with Hannah Lordan)
DCOM1 Group B with myself: Monday 12:00 in F1. 3 and Thursday 12:00 in B165
DNET1 with myself: Monday 14:00 in B248 and Wednesday 15:00 in C212
WEB1 with Ben O’Shaugnessy: Monday 16:00 in C212 and Thursday 15:00 in E7

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

2 comments

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October 29, 2015 at 5:22 pm

Student

Hi, we got the results back today and I did not do nearly as well as I hoped I would. This test was much easier than the last and I don’t know how I didn’t do better because I’m sure I’ve attended every lecture and I did study prior to this exam.

What questions did I get wrong?

October 29, 2015 at 5:47 pm

J.P. McCarthy

For question 3 you didn’t know that the principal is the starting amount. You calculated what €976.70 would amount to after 6 years at 4.5% interest. So you said:

Start (principal): €976.70
End: €1,271.92

while the question asked what principal would amount to €976.70:

Start (principal): ???
End: €976.70

At the very least your answer should have been less than €976.70. To do this properly you say let $P$ be the principal that amounts to €976.70. Therefore $P$ satisfies the equation:

$976.70=P(1+0.045)^6.$

Solving this equation for $P$ gives you the solution.

For question 4 you just did

$\displaystyle \frac{4\pi}{5}\approx 2.5$ …

What you had was

$\frac{4\pi}{5}\text{ rad}$ .

The way I recommended that ye do these was as follows. Know that

$\displaystyle \pi\text{ rad}=180^\circ\Rightarrow \text{rad}=\frac{180}{\pi}^\circ$ .

Therefore

$\frac{4\pi}{5}\text{ rad}=\frac{4\pi}{5}\frac{180}{\pi}^\circ\underset{\text{EC}}{=}144^\circ$ .

For question 5… I have no idea what you actually did wrong. If you wrote sohcahtoa you would note that

$\displaystyle \cos(40^\circ)=\frac{25}{X}$

and you could solve for $X$ .

For question 6 you assumed that all of the sides had the same length. This is only true for triangles where all angles are $60^\circ$ . In fairness you made a good effort after this but you know… MCQ. What you needed to do was use

$\displaystyle A=\frac12 bc\sin A$

on the top wedge of the triangle.

For question 6 you have an even number of numbers:

$8,9,9,11,14,16$ .

The median separates the numbers into big and small:

$8,9,9\,|\,11,14,16$ .

With an even number of terms you take the average of the two middle numbers so you should have gotten 10. You said 11.

For question 8… well question 8 was a little harder. Your attempt made no sense. Four of five numbers are 162, 183, 119 and 218. You calculated

$\displaystyle \frac{162+183+119+218}{5}$ .

I have no idea why. You either let $x$ be the fifth number and solve

$\displaystyle \frac{162+183+119+218+x}{5}=174$

Or else use the alternative method (see the top of page 72 of the notes).

For question 9. You started well with:

$V=\pi r^2h\Rightarrow 0.9425=\pi r^2 (1.2)$ .

After this you did something which makes no sense. You ‘moved’ some symbols around. Perhaps this question is tough because we are only properly doing equations now (although I did talk about these principles earlier in the year)… see in class now how we would deal with this.

$\displaystyle 0.9425=\pi r^2 h\Rightarrow 1.2\pi r^2=0.9425$

$\displaystyle \underset{\div 1.2\pi}{\Rightarrow} \frac{1.2\pi r^2}{1.2\pi }=\frac{0.9425}{1.2\pi}$

$\displaystyle \Rightarrow r^2\approx 0.25$

$\displaystyle \underset{\sqrt{\cdot}}{\Rightarrow} r\approx \pm\sqrt{0.25}=0.5$ m,

because $r$ is positive.

For question 10 you calculated the area when you were asked to calculate the perimeter.

For question 11. (i) you lost a half mark for not giving your answer correct to the nearest whole number. For part (ii) you failed completely to apply the definition of sine. What you should have had was

$\displaystyle \sin(\theta)=\frac{24}{50}$ ,

to which you can apply $\sin^{-1}$ to both sides. Instead you calculated

$\displaystyle \sin(24/50)$ … this is just sine of $0.48^\circ$ which has nothing to do with the question.

You lost a mark in question 11 because you found the area of the whole circle when there was only a semi-circle.

In question 13 — where you clearly put in work — you lost half a mark for not giving your answer correct to two decimal places.

That is twice you lost marks for not reading the questions properly.

Regards,
J.P.

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MATH6000: Week 6

Assessment 2 Results

Week 6

DCOM

WEB/DNET

Week 7

Tutorials

Study

Student Resources

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J.P. McCarthy Maths

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MATH6000: Week 6

Assessment 2 Results

Week 6

DCOM

WEB/DNET

Week 7

Tutorials

Study

Student Resources

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J.P. McCarthy Maths

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