Answer:
ai) n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) see attachment
bi) n(C⋂G) = x = 1
bii) n(G) only = 3
Step-by-step explanation:
Let chemistry = C
Economic = E
Government = G
n(E) = 12
n(G) = 8
n(C) = 7
ai) number of pupils for economics and chemistry = 0
number of pupils for economics and government = 4
The set notation for both:
n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) find attached the Venn diagram
bi) n(C⋂G) = ?
Let number of n(C⋂G) = x
From the Venn diagram
n(C) only = 12-4 = 8
n(G) only = 8-(4+x) = 4-x
n(E) only = 7-x
n(E⋂C⋂G) = 0
n(E⋂C) = 0
n(E⋂G) = 4
Total: 8+ 4-x + 7-x + x + 0+0+4 = 22
23 -x = 22
23-22 = x
x = 1
n(C⋂G) = x = 1
Number of pupils that take both chemistry and government = 1
(bii) government only = n(G) only = 4-x
n(G) only = 4-1 = 3
Number of students that take government only = 3
Okay! thank you so much for the awarness!
Answer:
x =19.20
Step-by-step explanation:
We can use ratios to solve
14 wings 12 wings
------------- = ---------------
22.40 x
Using cross products
14x = 22.40 * 12
14x =268.8
Divide each side by 14
14x/14 = 268.8/14
x =19.20
Sent a picture of the solution to the problem (s).
Answer:
b. left- skewed
Step-by-step explanation:
edge 2020