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## PowerPoint Slideshow about ' Main Grid' - felcia

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This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2004

The questions are sorted into topics based on the specific outcomes for the Intermediate 2 course.

To access a particular question from the main grid click on the question number.

To get the solution for a question

press the space bar.

To access the formula sheet press the button

To begin click on Main Grid button.

F

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START Page

Number Freq Cum Freq

4 3 3

5 7 10

6 2 12

7 3 15

8 1 16

9 2 18

10 2 20

(b) Prob(>7) = 5/20 = ¼

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Solution

Intercept = 1 (where line crosses y axis)

Gradient = 2

(measure from graph by vertical ÷ horizontal

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OTP = 90° – 24° = 66°

OQT = OTP (Isosceles)

TQP = 180° – 66° = 114°

QPT = 180° - (114° + 24°) = 42°

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Solution

Solution

Position of median, Q2 = (25 + 1) ÷ 2 = 13th number

Q2 = 50

Q1 = 49 (position 6/7th)

Q3 = 51.5 (position 19/20th)

No. of chocolates in a box

No. of chocolates

(c) Semi IQR = (Q3 – Q1) ÷ 2

= (51.5 – 49) ÷ 2 = 1.25

The second sample has a higher semi IQR

so more variation in the sample

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- (-2, -16)
- (6, -16)
- 2+8+4 = 14
- (14, -16)
- Equation of parabola is
- y = (x – 14)2 -16

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Solution

Solution

Solution

Solution

- Since all have grown by same amount the mean will be up
- + 4mm to 20.5mm. [ mean= (99 + 6 × 4)÷6 = 20.5]
- Spread is the same so standard deviation unchanged at 1.87.

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Solution

Solution

Solution

Solution

(b) 13x + 6y = 54.50 equ 2 x2

(c) Solving simultaneously

42x + 12y = 165

26x + 12y = 109

Subtract 16x = 56

x = 3.5

Sub into equ 1

14 × 3.5 + 4y = 55

4y = 55 – 49

4y = 6

y = 1.5

Check equ 2 13 × 3.5 + 6 × 1.5 = 54.50

Adult ticket is £3.50 and child ticket is £1.50

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Solution

Solution

Solution

x

2+x

A

2

B

(a) Area A = length x breadth = 2(2 + x) = 4 + 2x

Area B = 2x

Total area = A + B

= 4 + 2x + 2x

= 4 + 4x

(b) Area = 4 + 4x = 18

So 4 + 4x = 18

4x = 18 – 4

4x = 14

x = 14 ÷ 4 = 3.5m

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Solution

Solution

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