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
No.
The sum of the areas of the two smaller squares is 5^2 + 3^2 = 34
The area of the larger square is 8^2=64
34 does not = 64
"Energy equals mass times the speed of light squared." On the most basic level, the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing. Under the right conditions, energy can become mass, and vice versa.
(1/3)t = (2/3)c, where "t" is for teapot and "c" is for the container.
Multiply both sides by (3/2) in order to get 1 on the right-hand side.
(3/2)(1/3)t = (3/2)(2/3)C
simplify the equation to get the answer which is:
(1/2) teacup = 1 full container.
Hope this helps :)
We can tell from the data that there is a midpoint between the lowest and highest point of the clock, which is at a height of 9.5 feet.
Moreover, the lowest point occurs at 6 o clock, and the highest occurs at 12 o clock.
The amplitude of variation from the mid-point is 0.5 feet given by (10 - 9) / 2.
Finally, the time period for the equation is 12 hours. Thus, the answer is:
h = 0.5cos(πt/6) + 9.5, option B
