Answer
a) y | p(y)
25 | 0.8
100 | 0.15
300 | 0.05
E(y) = ∑ y . p(y)
E(y) = 25 × 0.8 + 100 × 0.15 + 300 × 0.05
E(y) = 50
average class size equal to E(y) = 50
b) y | p(y)
25 | 
100 | 
300 | 
E(y) = ∑ y . p(y)
E(y) = 25 × 0.4 + 100 × 0.3 + 300 × 0.3
E(y) = 130
average class size equal to E(y) = 130
c) Average Student in the class in a school = 50
Average student at the school has student = 130
Answer
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Step-by-step explanation:
Answer:
See explanation
Step-by-step explanation:
f(x) = -2lx-3| +1
1.lx-3|==>Translate the basic absolute value graph f(x)=|x| 3 units to the right
2. -lx-3| ==> Flip the graph over the x-axis
3. -2lx-3| ==> Double all the y-coordinates of the graph
4. -2lx-3| +1 ==> Translate the graph 1 unit up
<span>Let the width of the rectangular plot of land be 'x' yards.
Given that the length of the rectangular plot of land is 10 yards more than its width.
So, width of the rectangular plot of land = (x + 10) yards.
Also given that the area of the rectangular plot of land is 600 square yards.
We know that, area of a rectangle = length * width
That is, (x+10) * x = 600
x^2 + 10x = 600
x^2 + 10x - 600 = 0
x^2 + 30x - 20x -600 = 0
x(x + 30) - 20(x + 30) = 0
(x +30)(x -20) =0
Therefore, either (x + 30) = 0 or (x - 20) = 0
If x + 30 = 0, then x = -30 and
If x - 20 = 0, then x = 20
Since 'x' represents the width of a rectangular plot of land it cannot be negative.
Therefore,
width of the rectangular plot of land = 20 yards
length of the rectangular plot of land= x + 10 = 30 yards</span>
