Question
The owner of a farming store tracks the cost of bubble wrap for packing pictures like the one shown. How can you use the data to estimate the cost of the bubble wrap for a picture with a length of 75 in.
Hint:
Hint:-
1. When the difference between 2 consecutive differences for output values (y values) for a given constant change in the input values (x values) is constant. i.e. dy(n)- dy(n-1) is constant for any value of n, the function is known as a quadratic function.
2. Regression is a statistical tool used to find a model that can represent the relation between a given change in dependant variable (output values/ y values) for a given change in independent variable (input values/ x values).
Quadratic Equation using regression can be represented as-
Y = aX2 + bX + c, where-
Σy = nc + b(Σx) + a(Σx2)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
The correct answer is: Cost of the bubble wrap for a picture with a length of 75 in. will be approximately $ 8.9.
Step-by-step solution:-
From the given information, we get-
x coordinates in the given table pertains to length of bubble wrap (in inches) and y coordinates pertain to the cost of such bubble wrap.
Now, from the given table, we observe the following readings-
x1 = 6, y1 = 0.1;
x2 = 12, y2 = 0.31;
x3 = 18, y3 = 0.62;
x4 = 24, y4 = 1.04;
x5 = 30, y5 = 1.57.
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 12 - 6 = 6
dx2 = x3 - x2 = 18 - 12 = 6
dx3 = x4 - x3 = 24 - 18 = 6
dx4 = x5 - x4 = 30 - 24 = 6
Difference between 2 consecutive y values-
dy1 = y2 - y1 = 0.31 - 0.1 = 0.21
dy2 = y3 - y2 = 0.62 - 0.31 = 0.31
dy3 = y4 - y3 = 1.04 - 0.62 = 0.42
dy4 = y5 - y4 = 1.57 - 1.04 = 0.53
We observe that the difference for every consecutive x values is constant i.e. 1 but for y values the difference is not constant.
Hence, the given function is not a linear function.
b). Now, difference between 2 consecutive differences for y values-
dy2 - dy1 = 0.31 - 0.21 = 0.1
dy3 - dy2 = 0.42 - 0.31 = 0.11
dy4 - dy3 = 0.53 - 0.42 = 0.11
We observe that the difference of differences of 2 consecutive y values are almost constant i.e. 0.11.
Hence, the given function is a quadratic function.
Using Quadratic Regression formula and values from the adjacent table-
Y = aX2 + bX + c, where-
Σy = nc + b(Σx) + a(Σx2)
∴ 3.64 = 5c + b(90) + a(1,980)
∴ 3.64 = 5c + 90b + 1,980a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 87.54 = c(90) + b(1,980) + a(48,600)
∴ 87.54 = 90c + 1,980b + 48,600a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 2,261.16 = c(1,980) + b(48,600) + a(12,68,784)
∴ 2,261.16 = 1,980c + 48,600b + 12,68,784a ....................... (Equation iii)
Dividing Equation 2 by 18, we get-
2,700a + 110b + 5c = 4.86 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
2,700a + 110b + 5c = 4.86 …............................................... (Equation iv)
- 1,980a + 90b + 5c = 3.64 …............................................... (Equation i)
720a + 20b = 1.22 .................................................. (Equation v)
Multiplying Equation ii with 22, we get-
10,69,200a + 43,560b + 1,980c = 1,925.88 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
12,68,784a + 48,600b + 1,980c = 2,261.16 ......................... (Equation iii)
- 10,69,200a + 43,560b + 1,980c = 1,925.88 ......................... (Equation vi)
1,99,584a + 5,040b = 335.28 ......................... (Equation vii)
Multiplying Equation v with 252, we get-
1,81,440a + 5,040b = 307.44 ............................................... (Equation viii)
Subtracting Equation viii from Equation vii, we get-
1,99,584a + 5,040b = 335.28 ............................................... (Equation vii)
- 1,81,440a + 5,040b = 307.44 ............................................... (Equation viii)
18,144a = 27.84
i.e. 18,144a = 27.84
∴ a = 27.84/ 18,144 ................................... (Dividing both sides by 18,144)
∴ a = 0.0015
Substituting a = 0.0015 in Equation v, we get-
720a + 20b = 1.22 .................................................. (Equation v)
∴ 720(0.0015) + 20b = 1.22
∴ 1.08 + 20b = 1.22
∴ 20b = 1.22 - 1.08 ........................................ (Taking all constants together)
∴ 20b = 0.14
∴ b = 0.14/20 ............................................ (Dividing both sides by 20)
∴ b = 0.007
Substituting a = 0.0015 and b = 0.007 in Equation i, we get-
1980a + 90b + 5c = 3.64 .............................. (Equation i)
∴ 1,980 (0.0015) + 90 (0.007) + 5c = 3.64
∴ 2.97 + 0.63 + 5c = 3.64
∴ 3.60 + 5c = 3.64
∴ 5c = 3.64 - 3.60 ..................... (Taking all constants together)
∴ 5c = 0.04
∴ c = 0.04/5 ........................... (Dividing both sides by 5)
∴ c = 0.008
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = 0.0015 X2 + 0.007 X + 0.008
∴ for x = 75,
Y = 0.0015 (75)2 + 0.007 (75) + 0.008
∴ Y = 8.4375 + 0.525 + 0.008
∴ Y = 8.9705
Final Answer:-
∴ Cost of the bubble wrap for a picture with a length of 75 in. will be approximately $ 8.9.
x2 = 12, y2 = 0.31;
x3 = 18, y3 = 0.62;
x4 = 24, y4 = 1.04;
x5 = 30, y5 = 1.57.
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 12 - 6 = 6
dx2 = x3 - x2 = 18 - 12 = 6
dx3 = x4 - x3 = 24 - 18 = 6
dx4 = x5 - x4 = 30 - 24 = 6
Difference between 2 consecutive y values-
dy1 = y2 - y1 = 0.31 - 0.1 = 0.21
dy2 = y3 - y2 = 0.62 - 0.31 = 0.31
dy3 = y4 - y3 = 1.04 - 0.62 = 0.42
dy4 = y5 - y4 = 1.57 - 1.04 = 0.53
We observe that the difference for every consecutive x values is constant i.e. 1 but for y values the difference is not constant.
Hence, the given function is not a linear function.
b). Now, difference between 2 consecutive differences for y values-
dy2 - dy1 = 0.31 - 0.21 = 0.1
dy3 - dy2 = 0.42 - 0.31 = 0.11
dy4 - dy3 = 0.53 - 0.42 = 0.11
We observe that the difference of differences of 2 consecutive y values are almost constant i.e. 0.11.
Hence, the given function is a quadratic function.
Using Quadratic Regression formula and values from the adjacent table-
Y = aX2 + bX + c, where-
Σy = nc + b(Σx) + a(Σx2)
∴ 3.64 = 5c + b(90) + a(1,980)
∴ 3.64 = 5c + 90b + 1,980a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 87.54 = c(90) + b(1,980) + a(48,600)
∴ 87.54 = 90c + 1,980b + 48,600a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 2,261.16 = c(1,980) + b(48,600) + a(12,68,784)
∴ 2,261.16 = 1,980c + 48,600b + 12,68,784a ....................... (Equation iii)
Dividing Equation 2 by 18, we get-
2,700a + 110b + 5c = 4.86 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
2,700a + 110b + 5c = 4.86 …............................................... (Equation iv)
- 1,980a + 90b + 5c = 3.64 …............................................... (Equation i)
720a + 20b = 1.22 .................................................. (Equation v)
Multiplying Equation ii with 22, we get-
10,69,200a + 43,560b + 1,980c = 1,925.88 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
12,68,784a + 48,600b + 1,980c = 2,261.16 ......................... (Equation iii)
- 10,69,200a + 43,560b + 1,980c = 1,925.88 ......................... (Equation vi)
1,99,584a + 5,040b = 335.28 ......................... (Equation vii)
Multiplying Equation v with 252, we get-
1,81,440a + 5,040b = 307.44 ............................................... (Equation viii)
Subtracting Equation viii from Equation vii, we get-
1,99,584a + 5,040b = 335.28 ............................................... (Equation vii)
- 1,81,440a + 5,040b = 307.44 ............................................... (Equation viii)
18,144a = 27.84
i.e. 18,144a = 27.84
∴ a = 27.84/ 18,144 ................................... (Dividing both sides by 18,144)
∴ a = 0.0015
Substituting a = 0.0015 in Equation v, we get-
720a + 20b = 1.22 .................................................. (Equation v)
∴ 720(0.0015) + 20b = 1.22
∴ 1.08 + 20b = 1.22
∴ 20b = 1.22 - 1.08 ........................................ (Taking all constants together)
∴ 20b = 0.14
∴ b = 0.14/20 ............................................ (Dividing both sides by 20)
∴ b = 0.007
Substituting a = 0.0015 and b = 0.007 in Equation i, we get-
1980a + 90b + 5c = 3.64 .............................. (Equation i)
∴ 1,980 (0.0015) + 90 (0.007) + 5c = 3.64
∴ 2.97 + 0.63 + 5c = 3.64
∴ 3.60 + 5c = 3.64
∴ 5c = 3.64 - 3.60 ..................... (Taking all constants together)
∴ 5c = 0.04
∴ c = 0.04/5 ........................... (Dividing both sides by 5)
∴ c = 0.008
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = 0.0015 X2 + 0.007 X + 0.008
∴ for x = 75,
Y = 0.0015 (75)2 + 0.007 (75) + 0.008
∴ Y = 8.4375 + 0.525 + 0.008
∴ Y = 8.9705
Final Answer:-
∴ Cost of the bubble wrap for a picture with a length of 75 in. will be approximately $ 8.9.
Check for extraneous solutions
,where H is the height in centimeters and M is the mass in kilograms. A doctor calculates a particular dose of medicine for a patient whose BSA is less than 1.9. If the patient is 160 cm tall, what must the mass of the person be for the dose to be appropriate?
