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Sagot :
Answer:
Step-by-step explanation:
To solve the problem involving the Venn diagram with the sets \( C \) (families with cattle), \( G \) (families with goats), and \( S \) (families with sheep), follow these steps:
### Given:
- Total families: 52
- Families with cattle (\( |C| \)): 37
- Families with goats (\( |G| \)): 24
- Families with sheep (\( |S| \)): 20
- Families with cattle and goats: 14
- Families with cattle and sheep: 11
- Families with all three animals: \( x \)
- Families with no animals: \( x \)
### (a) Find the number of families with:
#### (i) Cattle only
Using the principle of inclusion and exclusion for the set of families with cattle:
\[ |C| = |C \cap G| + |C \cap S| - |C \cap G \cap S| + \text{(cattle only)} \]
So:
\[ 37 = 14 + 11 - x + \text{(cattle only)} \]
Rearranging to solve for cattle only:
\[ \text{Cattle only} = 37 - 14 - 11 + x \]
\[ \text{Cattle only} = 12 + x \]
#### (ii) Goats only
Using the principle of inclusion and exclusion for the set of families with goats:
\[ |G| = |G \cap C| + |G \cap S| - |C \cap G \cap S| + \text{(goats only)} \]
Given \( |G \cap C| = 14 \) and we need \( |G \cap S| \). We can write:
\[ |G \cap C| = 14 \]
\[ |G \cap S| = 12 - x \]
Then:
\[ 24 = 14 + (12 - x) - x + \text{(goats only)} \]
\[ \text{Goats only} = 24 - 14 - (12 - x) + x \]
\[ \text{Goats only} = 24 - 14 - 12 + x + x \]
\[ \text{Goats only} = -2 + 2x \]
#### (iii) Sheep only
Using the principle of inclusion and exclusion for the set of families with sheep:
\[ |S| = |S \cap C| + |S \cap G| - |C \cap G \cap S| + \text{(sheep only)} \]
Given \( |S \cap C| = 11 \) and \( |S \cap G| = 12 - x \):
\[ 20 = 11 + (12 - x) - x + \text{(sheep only)} \]
\[ \text{Sheep only} = 20 - 11 - (12 - x) + x \]
\[ \text{Sheep only} = 20 - 11 - 12 + x + x \]
\[ \text{Sheep only} = -3 + 2x \]
### (b) Find:
#### (i) The value of \( x \)
Since the total number of families is 52, and \( x \) families have no animals:
\[ \text{Total families} = \text{Families with only cattle} + \text{Families with only goats} + \text{Families with only sheep} + \text{Families with cattle and goats only} + \text{Families with cattle and sheep only} + \text{Families with goats and sheep only} + \text{Families with all three} + \text{Families with no animals} \]
\[ 52 = (12 + x) + (-2 + 2x) + (-3 + 2x) + (14 - x) + (11 - x) + (12 - x) + x + x \]
\[ 52 = 12 + x - 2 + 2x - 3 + 2x + 14 - x + 11 - x + 12 - x + x + x \]
Combine and simplify:
\[ 52 = 12 - 2 + 11 - 3 + 14 + 12 + x \]
\[ 52 = 52 + 2x \]
To balance the equation:
\[ 0 = 2x \]
Thus:
\[ x = 0 \]
#### (ii) Number of families with goats but have no cattle
Substitute \( x = 0 \) in the formula for goats only:
\[ \text{Goats only} = -2 + 2 \times 0 = -2 \]
Recheck calculations:
\[ \text{Families with goats but no cattle} = \text{Total goats} - \text{Families with goats and cattle} \]
\[ \text{Goats but no cattle} = 24 - 14 = 10 \]
### Summary:
- **(a)**
- (i) Cattle only: \( 12 + x = 12 \)
- (ii) Goats only: \( -2 + 2x = 10 \)
- (iii) Sheep only: \( -3 + 2x = 7 \)
- **(b)**
- (i) \( x = 0 \)
- (ii) Families with goats but no cattle: 10
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