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Bond [772]
3 years ago
5

The answer to all of these problems

Mathematics
1 answer:
timurjin [86]3 years ago
4 0
Its easier to ask somebody face-to-face dude
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How many numbers can you make with 01349
Alchen [17]

Answer:

The answer is five

Step-by-step explanation:

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2 years ago
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The data set represents a month-to-month progression of gasoline prices over the course of several months in an unspecified city
Tresset [83]

The quadratic regression equation for this data set is \rm y=-0.143x^2+0.59x+2.82\\.

<h3>What is the general form of a quadratic equation?</h3>

The general form of the quadratic equation is given by;

\rm ax^2+bx+c=0

Where; a, b, and c are the constants.

x;   0       1         2        3          4         5

y;  2.82  3. 29  3. 46  3. 33   2. 88  2. 24

From the table when the value of x = 0 the value of y is 2.82.

Then,

\rm y=ax^2+bx+c\\\\x=0\\\\2.82=a0^2+b(0)+c\\\\c=2.82

From the table when the value of x = 3 the value of y is 3.33.

\rm y=ax^2+bx+c\\\\x=3\\\\2.82=a(3)^2+b(3)+c\\\\9a+3b+2.82=2.82\\\\9a=-3b\\\\b=-3a\\\\

From the table when the value of x = 5 the value of y is 2.24.

\rm y=ax^2+bx+c\\\\x=3\\\\2.24=a(5)^2+(-3a)(5)+c\\\\25a-15a+2.82=2.24\\\\40a=2.24-2.82\\\\10a=-0.68\\\\a= \dfrac{-0.68}{10}\\\\a=-0.143

And the value of b is;

\rm b=3a\\\\b=3(-0.14)\\\\b=0.59

Therefore,

The quadratic regression equation for this data set is;

\rm y=ax^2+bx+c\\\\y=-0.143x^2+0.59x+2.82\\\\

Hence, the quadratic regression equation for this data set is \rm y=-0.143x^2+0.59x+2.82\\.

To know more about the Quadratic equation click the link given below.

brainly.com/question/14935613

8 0
2 years ago
Convert 2/5
netineya [11]
The answer is b) 40%
8 0
3 years ago
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Matt bought a 20 pound bag of dog food for his dog, Hunter. On Monday Hunter ate 1 <img src="https://tex.z-dn.net/?f=%20%5Cfrac%
Ahat [919]
You need to add 1 3/4, 2 1/3, and 2 3/4. But before you do that you need to make them decimals. Add 1.75, 2.33, and 2.75 Equals 6.83 Then subtract it by 20.00 Equals 13.17 Then you make that a decimal 13 17/100
3 0
3 years ago
Find the derivative of the function y = 2x^2 - 13x + 5 and use it to find the equation of the line tangeant to the curve at x =
ser-zykov [4K]
Y = 2x^2 - 13x + 5
dy/dx = 4x - 13

At x = 3, dy/dx = 4(3) - 13 = 12 - 13 = -1
and y = 2(3)^2 - 13(3) + 5 = 2(9) - 39 + 5 = 18 - 34 = -16

The tangent line at x = 3, passes through the point (3, -16) and has a slope of -1
y - (-16) = -1(x - 3)
y + 16 = -x + 3
y = -x - 13 is the required equation.
3 0
2 years ago
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