Stanford University Research: The most important aspect of a student’s ideal relationship with mathematics

Source: Taken from Research by Stanford, Education: EDUC115N How to Learn Math

This word cloud was generated on August 9th based on 850 responses to the prompt “Please submit a word that, in your opinion, describes the most important aspect of a student’s ideal relationship with mathematics.”

H2 Maths A Level 2012 Solution, Paper 2 Q5; H2 Maths Tuition

5(i)(a)

$P(\text{patient has the disease and test positive})=0.001(0.995)=9.95\times 10^{-4}$

$P(\text{patient does not have the disease and he tests positive})=(1-0.001)(1-0.995)=4.995\times 10^{-3}$

$P(\text{result of the test is positive})=9.95\times 10^{-4}+4.995\times 10^{-3}=5.99\times 10^{-3}$

(b)

Let A=patient has disease

Let B=result of test is positive

$\displaystyle\begin{array}{rcl}P(A|B)&=&\frac{P(A\cap B)}{P(B)}\\ &=&\frac{(0.001)(0.995)}{5.99\times 10^{-3}}\\ &=&0.166 \end{array}$

Note that the probability is surprisingly quite low! (This is called the False positive paradox, a statistical result where false positive tests are more probable than true positive tests, occurring when the overall population has a low incidence of a condition and the incidence rate is lower than the false positive rate. See http://en.wikipedia.org/wiki/False_positive_paradox)

(ii)

$\displaystyle P(A|B)=\frac{(0.001)p}{(0.001)p+(1-0.001)(1-p)}=0.75$

By GC, $p=0.999666$ (6 d.p.)

A Level H2 Maths 2012 Paper 2 Q3 Solution; H2 Maths Tuition

A Level H2 Maths 2012 Paper 2 Q3 Solution

(i)

(The graph above is drawn using the Geogebra software 🙂 )

(ii)

$x^3+x^2-2x-4=4$

$x^3+x^2-2x-8=0$

By GC, $x=2$

By long division, $x^3+x^2-2x-8=(x-2)(x^2+3x+4)$

The discriminant of $x^2+3x+4$ is

$D=b^2-4ac=3^2-4(1)(4)=-7<0$

Hence, there are no other real solutions (proven).

(iii) $x+3=2$

$x=-1$

(iv)

(v)

$|x^3+x^2-2x-4|=4$

$x^3+x^2-2x-4=4$ or $x^3+x^2-2x-4=-4$

$x^3+x^2-2x-8=0$ or $x^3+x^2-2x=0$

$x^3+x^2-2x-8=0 \implies x=2$ (from part ii)

$x^3+x^2-2x=x(x^2+x-2)=x(x-1)(x+2)=0$

$x=0,1,-2$

In summary, the roots are $-2,0,1,2$

2012 H2 Maths Prelim Solution: SRJC/II/8(iv)

8(iv)

Using the model $\displaystyle y=a+\frac{b}{x-2}$ , estimate the total fertility rate for a particular country Z when its GDP per capita is USD$1000, giving your answer to 1 decimal place and comment on the reliability of the estimate.

First, we need to remove the outlier (40,6.6) as mentioned in part iii.

Then, performing linear regression with GC, (with variables $\frac{1}{x-2}$ , $y$ ), we get:

$a=0.974686$

$b=6.86442$

Substituting $x=1$ , we get $\displaystyle y=a+\frac{b}{1-2}=-5.9$ (1 d.p.)

Since we cannot have a negative fertility rate (the average number of children that would be born to a woman ), the estimate obtained for $y$ is not reliable.

MI H2 Maths Prelim Solutions 2010 Paper 2 Q7 (P&C)

MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)

Question:
7) Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if
(i) there is no restriction, [1]
(ii) the men and women are seated alternately, [2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3]
(iv) members of the same family are seated together and the seats are numbered. [2]

Solution:

(i) (10-1)!=9!=362880

(ii) First fix the men’s sitting arrangement: (5-1)!

Then the remaining five women’s total number of arrangements are: 5!

Total=4! x 5!=2880

(iii) Fix the 2 families (as a group) and the 2 men: (4-1)! x 3! x 3!

(3! to permute each family)

By drawing a diagram, the two women have 4 slots to choose from, where order matters: $^4 P_2$

Total = $(4-1)! \times 3! \times 3! \times ^4 P_2 = 2592$

(iv)

We first find the required number of ways by treating the seats as unnumbered: $(6-1)!\times 3!\times 3! =4320$

Since the seats are numbered, there are 10 choices for the point of reference, thus no. of ways = $4320 \times 10 =43200$