Answer:
2. The correct option is;
D. 112
3. The predicted number of cellular subscription per 100 people in Australia in 2030 is 217 subscriptions
4. The equation of the quadratic regression curve is y = x² - 6×x + 4.1
5. The closest option is;
C. 4191
Explanation:
2. The equation for the number of customers in a store is presented as follows;
y = -8.74·x² + 50.57·x + 39.02
Where:
x = Number of hours after opening
∴ When x = 3 hours
y = -8.74×3² + 50.57×3 + 39.02 = 112.07 ≈ 112 people
The correct option is D. 112
3. The expression of the regression equation is presented as follows
y = -0.107·x² + 9.008·x +28.851
Where:
x = Number of years since 1990, hence
When x = 2030 - 1990 = 40 Years
Hence, the predicted number of cellular subscriptions per people in 2030 is presented as follows;
y = -0.107×40² + 9.008×40 +28.851 = 217.971 Subscribers
To round down to the nearest whole number as y = 217 subscribers
4. The general form of a quadratic equation is presented as follows;
y = a·x² + b·x + c
When x = 0, y = 4.1, therefore;
4.1 = a×0² + b×0 + c = c
∴ c = 4.1
When x = 1, y = -0.9, therefore;
-0.9 = a×(1)² + b×1 + c = c
-0.9 = a + b + 4.1
∴ a + b = -0.9 - 4.1 = -5.0
When x = 2, y = -3.9, therefore;
-3.9 = a×(2)² + b×2 + c = c
-3.9 = 4·a + 2·b + 4.1
∴ 4·a + 2·b = -3.9 - 4.1 = -8.0
Thus we have two equations;
a + b = -5.0..................(1) and
4·a + 2·b = -8.0 ..........(2)
Multiply equation (1) by 2 and subtract it from equation (2), we have
4·a + 2·b - 2×(a + b) = *8.0 - (2 ×-5.0)
∴ 2·a = 2
a = 1
From equation (1,) we have;
a + b = -5.0..................(1)
Therefore, where a = 1 we have, 1 + b = -5.0
Hence, b = -5.0 - 1 = -6.0
Therefore, the equation of the quadratic regression curve is presented as follows;
y = x² - 6×x + 4.1
5. The quadratic regression equation is found as follows;
The general form of a quadratic equation is presented as follows;
y = a·x² + b·x + c
When x = 55, y = 3650, therefore;
3650= a×55² + b×55+ c................(1)
When x = 65, y = 4569, therefore;
4569= a×65² + b×65+ c ...........(2)
When x = 75, y = 6147, therefore;
6147= a×75² + b×75+ c............(3)
Solving the system of equation;
3650= a×55² + b×55+ c................(1)
4569= a×65² + b×65+ c ...........(2)
6147= a×75² + b×75+ c............(3)
Subtracting equation (1) from (2), we obtain;
4569 - 3650 = a×65² + b×65 + c - (a×55² + b×55 + c)
919 = 1200·a + 10·b...........(4)
Subtracting equation (2) from (3), we obtain;
6147 - 4569 = a×75² + b×75 + c - (a×65² + b×65+ c)
Which gives;
1578 = 1400·a + 10·b...........(5)
Subtracting equation (4) from equation (3), we obtain;
1578 - 919 = 1400·a + 10·b - (1200·a + 10·b)
659 = 200·a
a = 659/200= 3.295
Substituting the value of a in equation (4), we have;
919 = 1200×3.295 + 10·b = 3954 + 10·b
∴ 10·b = 919 -3954 = -3035
b = -3035/10 = -303.5
Substituting the value of a and b in equation (1), we have;
3650 = (3.295)×55² + (-303.5)×55+ c
3650 = -6725.125 + c
∴ c = 3650 - (- 6725.125)= 10375.125
Therefore, the quadratic regression equation is presented as follows;
y = 3.295·x - 303.5·b +10375.125
Hence, in 1961, x = 1961 - 1900 = 61, we have
y = 3.295×61² - 303.5×61 10375.125= 4122.32
Therefore, the closest option is C. 4191.