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Elanso [62]
3 years ago
11

The data set represents a bimonthly progression of gasoline prices over the course of several months in an unspecified city. Use

a graphing calculator to determine the quadratic regression equation for this data set.
x
0
2
4
6
8
y
2.86
2.89
2.93
3.04
3.11
Mathematics
1 answer:
hodyreva [135]3 years ago
4 0
Plugging in the data points into a quadratic regression application, we get the following regression:

y = ax^2 + bx + c,
where:
a= 0.00267857
b= 0.0110714
c= 2.85743
with an R² value of 0.9851
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To calculate a person’s BMI, you divide their weight( in kg) by their height ( in square meters). First convert 169 pounds to kg, which is 76.66. Then convert 65 inches to meters. 65 inches= 1.65 meters. Then divide 76.66/1.65= 46.46 kg/m
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A rectangle has a width of 59 centimeters and a perimeter of 238 centimeters. What is the rectangle's length?
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Answer:

60 cm

Step-by-step explanation:

2x the width and 2x the length gives you the perimeter. This can be represented by the equation:

2w + 2L = P

Use the information given in the word problem to fill in the variables.

2(59) + 2L = 238

118 + 2L = 238

move 118 to the other side of the equation.

2L = 238 - 118

2L = 120

move 2 to the other side of the equation.

L = \frac{120}{2}

L = 60

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What is the simplified form of (5x^3 + 2)(5x^3 -2)
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Simplify the expression.

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The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon
Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by x=3.24t-4.20t^2+1.07t^3 where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

x(t)=3.24t-4.20t^2+1.07t^3

x(1)=3.24(1)-4.20(1)^2+1.07(1)^3

x(1)=3.24-4.20+1.07

x(1)=0.11

b) At t= 2 s

x(t)=3.24t-4.20t^2+1.07t^3

x(2)=3.24(2)-4.20(2)^2+1.07(2)^3

x(2)=6.48-16.8+8.56

x(2)=-1.76

c) At t= 3 s

x(t)=3.24t-4.20t^2+1.07t^3

x(3)=3.24(3)-4.20(3)^2+1.07(3)^3

x(3)=9.72-37.8+28.89

x(3)=0.81

d) At t= 4 s

x(t)=3.24t-4.20t^2+1.07t^3

x(4)=3.24(4)-4.20(4)^2+1.07(4)^3

x(4)=12.96-67.2+68.48

x(4)=14.24

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

\triangle x=x(4)-x(0)

\triangle x=14.24-0

\triangle x=14.24

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity  is given by,

\triangle x=x(4)-x(2)

\triangle x=14.24-(-1.76)

\triangle x=14.24+1.76

\triangle x=16

4 0
3 years ago
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