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Sagot :
Let’s go through the problem step by step and provide a detailed solution.
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### Q1:
Determine the number of pupils observed (noted as 'n'):
1. Let:
- [tex]\( R \)[/tex] = number of pupils who ate rice.
- [tex]\( Y \)[/tex] = number of pupils who ate yam.
- [tex]\( N \)[/tex] = number of pupils who ate nice.
2. Given:
- [tex]\( R + Y + N - 2 \)[/tex] (for pupils who ate both yam and nice) equals 7.
- It's stated that half of the observed pupils ate rice, hence [tex]\( R = \frac{1}{2}n \)[/tex].
- [tex]\( Y = 7\)[/tex] (since it’s explicitly stated).
3. Plugging these into the equation:
[tex]\[ \frac{1}{2}n + 7 + N - 2 = 7 \][/tex]
4. Simplify:
[tex]\[ \frac{1}{2}n + 5 + N = 7 \][/tex]
Given [tex]\( N \)[/tex] is related to the total [tex]\( n \)[/tex]:
5. Solve for [tex]\( n \)[/tex]:
[tex]\[ \frac{1}{2}n + 5 = 7 \][/tex]
[tex]\[ \frac{1}{2}n = 2 \][/tex]
[tex]\[ n = 4 \][/tex]
So, the total number of observed pupils is 4.
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### Q2:
(a) Find the rule of the mapping:
Given the table:
[tex]\[ \begin{tabular}{llllll} $x$ & 0 & 1 & 2 & 3 & 4 \\ $y$ & 2 & 5 & 8 & 11 & 14 \end{tabular} \][/tex]
Observing the pattern:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex].
- When [tex]\( x = 1 \)[/tex], [tex]\( y = 5 \)[/tex].
- When [tex]\( x = 2 \)[/tex], [tex]\( y = 8 \)[/tex].
The pattern indicates a linear relationship where the difference in consecutive y-values is 3. Thus, the equation can be written as:
[tex]\[ y = 3x + 2 \][/tex]
(b) Find the probability of getting totals with dice:
- (i) Less than 5:
The possible sums less than 5 when two dice (each having faces from 1 to 6) are rolled are: (1,1), (1,2), and (2,1), which make up 3 successful outcomes.
Total possible outcomes when two dice are rolled is [tex]\( 6 \times 6 = 36 \)[/tex].
[tex]\[ P(\text{less than 5}) = \frac{3}{36} = \frac{1}{12} \][/tex]
- (ii) Exactly 7:
The possible sums of 7 when two dice are rolled are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), which make up 6 successful outcomes.
Total possible outcomes = 36.
[tex]\[ P(\text{exactly 7}) = \frac{6}{36} = \frac{1}{6} \][/tex]
(c) Simplify the expression:
[tex]\[ \frac{3x}{(x-y)^2} \div \frac{x+y}{x-y} \][/tex]
First, rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{3x}{(x-y)^2} \times \frac{x-y}{x+y} \][/tex]
Cancel out the [tex]\((x-y)\)[/tex] term:
[tex]\[ \frac{3x}{(x-y)} \times \frac{1}{(x+y)} \][/tex]
Final simplified form:
[tex]\[ \frac{3x}{(x-y) \times (x+y)} \][/tex]
The more simplified result is:
[tex]\[ \frac{3x}{(x - y) \cdot (x + y)} \][/tex]
---
### Q1:
Determine the number of pupils observed (noted as 'n'):
1. Let:
- [tex]\( R \)[/tex] = number of pupils who ate rice.
- [tex]\( Y \)[/tex] = number of pupils who ate yam.
- [tex]\( N \)[/tex] = number of pupils who ate nice.
2. Given:
- [tex]\( R + Y + N - 2 \)[/tex] (for pupils who ate both yam and nice) equals 7.
- It's stated that half of the observed pupils ate rice, hence [tex]\( R = \frac{1}{2}n \)[/tex].
- [tex]\( Y = 7\)[/tex] (since it’s explicitly stated).
3. Plugging these into the equation:
[tex]\[ \frac{1}{2}n + 7 + N - 2 = 7 \][/tex]
4. Simplify:
[tex]\[ \frac{1}{2}n + 5 + N = 7 \][/tex]
Given [tex]\( N \)[/tex] is related to the total [tex]\( n \)[/tex]:
5. Solve for [tex]\( n \)[/tex]:
[tex]\[ \frac{1}{2}n + 5 = 7 \][/tex]
[tex]\[ \frac{1}{2}n = 2 \][/tex]
[tex]\[ n = 4 \][/tex]
So, the total number of observed pupils is 4.
---
### Q2:
(a) Find the rule of the mapping:
Given the table:
[tex]\[ \begin{tabular}{llllll} $x$ & 0 & 1 & 2 & 3 & 4 \\ $y$ & 2 & 5 & 8 & 11 & 14 \end{tabular} \][/tex]
Observing the pattern:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex].
- When [tex]\( x = 1 \)[/tex], [tex]\( y = 5 \)[/tex].
- When [tex]\( x = 2 \)[/tex], [tex]\( y = 8 \)[/tex].
The pattern indicates a linear relationship where the difference in consecutive y-values is 3. Thus, the equation can be written as:
[tex]\[ y = 3x + 2 \][/tex]
(b) Find the probability of getting totals with dice:
- (i) Less than 5:
The possible sums less than 5 when two dice (each having faces from 1 to 6) are rolled are: (1,1), (1,2), and (2,1), which make up 3 successful outcomes.
Total possible outcomes when two dice are rolled is [tex]\( 6 \times 6 = 36 \)[/tex].
[tex]\[ P(\text{less than 5}) = \frac{3}{36} = \frac{1}{12} \][/tex]
- (ii) Exactly 7:
The possible sums of 7 when two dice are rolled are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), which make up 6 successful outcomes.
Total possible outcomes = 36.
[tex]\[ P(\text{exactly 7}) = \frac{6}{36} = \frac{1}{6} \][/tex]
(c) Simplify the expression:
[tex]\[ \frac{3x}{(x-y)^2} \div \frac{x+y}{x-y} \][/tex]
First, rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{3x}{(x-y)^2} \times \frac{x-y}{x+y} \][/tex]
Cancel out the [tex]\((x-y)\)[/tex] term:
[tex]\[ \frac{3x}{(x-y)} \times \frac{1}{(x+y)} \][/tex]
Final simplified form:
[tex]\[ \frac{3x}{(x-y) \times (x+y)} \][/tex]
The more simplified result is:
[tex]\[ \frac{3x}{(x - y) \cdot (x + y)} \][/tex]
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