Intersection (∩) are elements that can be found in both sets

P = {-2, 0, 2, 4, 6}

Q = {-3, -1, 0, 2, 3, 5}

P ∩ Q = {0, 2}

P = {x: 1 ≤ x ≤ 6}

Read as x is such that 1 is less than or equal to x and x is less than or equal to 6.

P = {1, 2, 3, 4, 5, 6}

Q = {x: 2 < x < 9}

Read as x is such that 2 is less than x and x is less than 9.

Q = {3, 4, 5, 6, 7, 8}

Intersection (∩) are elements that can be found in both sets

(P ∩ Q) = {3, 4, 5, 6}

(P ∩ Q) = {x: 2 < x ≤ 6}

Note: ≤ means less than or equal to, hence the number at the right of it is inclusive.

Even numbers are number divisible by 2 without a remainder.

Set of even integers between 11 and 97 = {12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,80,92,94,96}

The total number of elements in the set = 43

Alternate Solution

Find the interval between 97 and 11 and divide by 2

97 - 11 = 86

Even integers between 11 and 97 = $\frac{86}{2}$ = 43

P = {-3 < x < 1}, read as -3 is less than x and x is less than 1

P = {-2, -1, 0}

Q = {-1 < x <3}, read as -1 is less than x and x is less than 3.

Q = {0, 1, 2)

Intersection (∩) are elements that can be found in both sets

P ∩ Q = {0}

P ∩ Q = {-1 < x < 1}, read as -1 is less than x and x is less than 1 which is true for the 0. -1 is less than 0 and 0 is also less than 1.

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

x:x < 7 is read as x is such that x is less than 7.

X = {1, 2, 3, 4, 5, 6}

Note: since X is a subset of the universal set (µ), members of X should be found in the universal set (µ)

y:y is a factor of 24 is read as y is such that y is a factor of 24.

Y = {1, 2, 3, 4, 6, 8}

Note: since Y is a subset of the universal set (µ), members of Y should be found in the universal set (µ)

Intersection (∩) are elements that can be found in both sets

X ∩ Y = {1, 2, 3, 4, 6}

µ = {x : 0 < x ≤ 10}

x : 0 < x ≤ 10 read as x is such that 0 is less than x and x is also less than or equal to 10

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

T = {prime numbers}

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Note: 1 is not a prime number.

T = {1, 2, 3, 5, 7}

Note: the elements of T must be found in the universal set µ since it is a subset.

T^' (T compliment) are list of elements found in the universal set but not in the set T

T^' = {1, 4, 6, 8, 9, 10}

M = {odd numbers}

Odd numbers are numbers not divisible by 2. Thus there is a remainder when divided by 2.

M ={1, 3, 5, 7, 9}

Note: the elements of M must be found in the universal set µ since it is a subset.

M^' (M compliment) are list of elements found in the universal set but not in the set M

M^' = {2, 4, 6, 8, 10}

Intersection (∩) are elements that can be found in both sets

T^' = {1, 4, 6, 8, 9, 10}

M^' = {2, 4, 6, 8, 10}

T^' ∩ M^' are the elements that can be found in both T^' and M^'

T^' ∩ M^' = {4, 6, 8, 10}

Prime number: A number that is divisible only by itself and 1

Note: 1 is not a prime number

Integer: A whole number (not a fraction) that can be positive, negative or zero. e.g ...,-5,-4,-3,-2,-1,0,1,2,3...

P is a set of positive (No negative) integers and less than 7

Rational Number: A number such as 5/8 that can be expressed as quotient or fraction $\frac{\mathrm{p}}{\mathrm{q}}$ of two integers.

Real Number: Any positive or negative number. This includes all integers and all rational and irrational numbers.

x:x is read as x is such that x is ...

Intersection (∩) are elements that can be found in both sets

P = {-2, 0, 2, 4, 6}

Q = {-3, -1, 0, 2, 3, 5}

P ∩ Q = {0, 2}

U = {Students}

n(U) = 120

M = {Students who passed Mathematics}

n(M) = 90

S = {Students who passed Science}

n(S) = 40

Let x = number of students who passed both subjects.

Covering Mathematics and solving for x.

90 + 40 - x + 5 = 120

135 - x = 120

135 - 120 = x

15 = x

Science only = 40 - x

Science only = 40 - 15

Science only = 25

P = {x: 1 ≤ x ≤ 6}

Read as x is such that 1 is less than or equal to x and x is less than or equal to 6.

P = {1, 2, 3, 4, 5, 6}

Q = {x: 2 < x < 9}

Read as x is such that 2 is less than x and x is less than 9.

Q = {3, 4, 5, 6, 7, 8}

Intersection (∩) are elements that can be found in both sets

(P ∩ Q) = {3, 4, 5, 6}

(P ∩ Q) = {x: 2 < x ≤ 6}

Note: ≤ means less than or equal to, hence the number at the right of it is inclusive.

(3 + x)(1 - x) > 9 - x²

Expand the bracket

3(1 - x) + x(1 - x) > 9 - x²

3 - 3x + x - x² > 9 - x²

3 - 2x - x² > 9 - x²

- x² cancels each other at both sides of the equation.

3 - 2x > 9
- 2x > 9 - 3
- 2x > 6

Divide both sides by - 2. When dividing by a negative, the sign changes direction.

x < $\frac{6}{-2}$

x < -3

Let M represent Mathematics
B represent Biology
x represent students that take only Biology

13 + 5 + x + 8 = 30
26 + x = 30
x = 30 - 26
x = 4

Number of students that take Biology = 5 + 4 = 9

Range = Largest Value - Smallest Value

Range = 83 - 47

Range = 36

P = {-3 < x < 1}, read as -3 is less than x and x is less than 1

P = {-2, -1, 0}

Q = {-1 < x <3}, read as -1 is less than x and x is less than 3.

Q = {0, 1, 2)

Intersection (∩) are elements that can be found in both sets

P ∩ Q = {0}

P ∩ Q = {-1 < x < 1}, read as -1 is less than x and x is less than 1 which is true for the 0. -1 is less than 0 and 0 is also less than 1.

Mean/Average = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Sum of the fractions = (1 + 2 + 3 + 4 + 5)[$\frac{1}{2}$ + $\frac{2}{3}$ + $\frac{3}{4}$ + $\frac{4}{5}$ + $\frac{5}{6}$ ]

The L.C.M of 2, 3, 4, 5 and 6 is 60.

Sum of the fractions = 15[$\frac{\mathrm{30x\; 1\; +\; 20\; x\; 2\; +\; 15\; x\; 3\; +\; 12\; x\; 4\; +\; 10\; x\; 5}}{60}$]

Sum of the fractions = 15[$\frac{\mathrm{30\; +\; 40\; +\; 45\; +\; 48\; +\; 50}}{60}$]

Sum of the fractions = 15[$\frac{213}{60}$]

Sum of the fractions = 15$\frac{213}{60}$ = $\frac{\mathrm{15\; x\; 60\; +\; 213}}{60}$ = $\frac{1113}{60}$

Number of numbers = 5

Mean = $\frac{1113}{60}$ ÷ 5

Every number is being divided by 1 but not written.

5 = $\frac{5}{1}$

Mean = $\frac{1113}{60}$ ÷ $\frac{5}{1}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

Mean = $\frac{1113}{60}$ x $\frac{1}{5}$

Mean = $\frac{\mathrm{1113\; x\; 1}}{\mathrm{60\; x\; 5}}$

Mean = $\frac{1113}{300}$ = 3.71

Arrange the numbers in ascending order.

6, 7, 7, 7, 8, 8, 8, 9, 9, 11

First Quartile (ungrouped data) = $\frac{\mathrm{n}}{4}$^th value

Where n is the number of items.

n = 10

First Quartile (ungrouped data) = $\frac{10}{4}$ = 2.5^th value

First Quartile = $\frac{\mathrm{7\; +\; 7}}{2}$ = 7.0

Mean = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{6030}{37}$ = 162.97297 ≈ 163.0 (one decimal place)

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

x:x < 7 is read as x is such that x is less than 7.

X = {1, 2, 3, 4, 5, 6}

Note: since X is a subset of the universal set (µ), members of X should be found in the universal set (µ)

y:y is a factor of 24 is read as y is such that y is a factor of 24.

Y = {1, 2, 3, 4, 6, 8}

Note: since Y is a subset of the universal set (µ), members of Y should be found in the universal set (µ)

Intersection (∩) are elements that can be found in both sets

X ∩ Y = {1, 2, 3, 4, 6}

Upper quartile - Lower quartile (25th percentile) = Interquartile range

Lower quartile (25th percentile) = 16

Interquartile range = 7

Upper quartile - 16 = 7
Upper quartile = 7 + 16
Upper quartile = 23

µ = {x : 0 < x ≤ 10}

x : 0 < x ≤ 10 read as x is such that 0 is less than x and x is also less than or equal to 10

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

T = {prime numbers}

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Note: 1 is not a prime number.

T = {1, 2, 3, 5, 7}

Note: the elements of T must be found in the universal set µ since it is a subset.

T^' (T compliment) are list of elements found in the universal set but not in the set T

T^' = {1, 4, 6, 8, 9, 10}

M = {odd numbers}

Odd numbers are numbers not divisible by 2. Thus there is a remainder when divided by 2.

M ={1, 3, 5, 7, 9}

Note: the elements of M must be found in the universal set µ since it is a subset.

M^' (M compliment) are list of elements found in the universal set but not in the set M

M^' = {2, 4, 6, 8, 10}

Intersection (∩) are elements that can be found in both sets

T^' = {1, 4, 6, 8, 9, 10}

M^' = {2, 4, 6, 8, 10}

T^' ∩ M^' are the elements that can be found in both T^' and M^'

T^' ∩ M^' = {4, 6, 8, 10}