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mr Goodwill [35]
3 years ago
11

Yo I know ya'll can see this question. Please help me out. I can't fail this.

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

  D.  y = 12.11·1.36^x . . . . . best <em>exponential</em> model

  A.  y = 2.49x^2 +7.29x +3.57 . . . . . best model of the choices offered

Step-by-step explanation:

None of the offered choices is very good, and the best of them is only the best by a relatively small amount. The exponential models cannot have been developed using any reasonable approach to model building. So, you have to figure out how to assess these models when none of them has much relation to the given data.

The horizontal asymptote of an exponential function is zero. Here, the values seem to have a horizontal asymptote of about 6. Consequently, you would look at the first few numbers and expect a vertical offset to the exponential function of about 6. Subtracting that from the remaining numbers, you might toss "13" as an outlier and use some of the others to compute the base of the exponent. If you do that using the points (0, 8) and (4, 88), you would compute the base to be ...

   ((88 -6)/(8 -6))^(1/(4-0)) ≈ 2.53

The multiplier of the exponential part is its value when x=0. We've estimated that to be 8-6 = 2. This gives an exponential model that looks like ...

  y = 6 + 2·2.53^x

____

<em>Comparison of our estimated model to the offered choices</em>

Selection A is a <em>quadratic</em> model. It has precisely the coefficients that are calculated by a spreadsheet or graphing calculator for a quadratic model. Though it gives the best fit of any of the offered choices, it is not the required <em>exponential</em> regression equation.

The base of selection C is about 5 times as great as needed, so it will vastly overestimate any points for x ≥ 1. It gives by far the worst fit of all of these choices.

The choice with the largest reasonable base is B, but it gives values for y that are less than any of those listed. As a consequence, its error is higher than necessary.

The remaining choice D gives a curve that is less than data points at the ends of the table and greater than the data values in the middle of the table. Of the offered exponential models, it has the least overall error. (You might pick choice D simply because it uses the same numbers as in choice C, but puts those numbers in places that make the function a better fit.)

_____

<em>Graphing calculator results</em>

If you ask a graphing calculator to give you an exponential model for these data, you can get either of ...

  •   y = 6.2 + 1.2·2.87^x . . . . . exponential with vertical offset
  •   y = 3.38·2.25^x . . . . no vertical offset; can't match x<0 very well

The first of these has about 1/10 the error of the last of these. Both have less than 1/10 the error of the available answer choices.

_____

<em>Comment on the attachment</em>

The black and red curves correspond to the first two answer choices. The functions associated with the other colors are shown at the left. The green points are those given in the problem statement.

The "total" numbers are the total squared error of each of the functions. Smaller errors mean the function is a better fit to the data.

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Tems11 [23]

Step-by-step explanation:

Since we have a vertical major axis, our ellipse is vertical.

The equation of a vertical ellipse is

\frac{(y - k) {}^{2} }{ {a}^{2} }  +  \frac{(x - h) {}^{2} }{ {b}^{2} }  = 1

where

(h,k) is the center

a is the semi major axis,

b is the semi minor axis

First, let plug in our center

\frac{(y + 3) {}^{2} }{ {a}^{2} }  + \frac{(x + 2) {}^{2} }{ {b}^{2} }  = 1

Semi means half, so

a is half of 14 which is 7

B is half of 8, which is 4.

\frac{(y + 3) {}^{2} }{49}  +  \frac{(x + 2) {}^{2} }{16}  = 1

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2 years ago
What's the equation of the line that's a perpendicular bisector of the segment connecting C (6, –12) and D (10, –8)? answers: A)
vitfil [10]

Answer:

A

Step-by-step explanation:

Perpendicular bisector of a line divides the line into 2 equal parts and it is perpendicular to the line.

First let's find the midpoint of CD. The point is where the perpendicular bisector will cut through the line.

midpoint= ( \frac{x1 + x2}{2} ,  \frac{y1 + y2}{2} )

Thus, midpoint of CD

= ( \frac{6 + 10}{2} , \frac{ - 12 - 8}{2} ) \\  = ( \frac{16}{2} , \frac{ - 20}{2} ) \\  = (8, - 10)

Gradient of line CD

=  \frac{y1 - y2}{x1 - x2}  \\  =  \frac{ - 12 -  ( - 8)}{6 - 10}  \\  =  \frac{ - 12 + 8}{ - 4} \\  =  \frac{ - 4}{ - 4}  \\  = 1

The product of the gradients of perpendicular lines is -1.

gradient if perpendicular bisector(1)= -1

gradient of perpendicular bisector= -1

y=mx +c, where m is the gradient and c is the y-intercept.

y= -x +c

Subst a coordinate to find c.

<em>Since the perpendicular bisector passes through the point (8, -10):</em>

When x=8, y= -10,

-10= -8 +c

c= -10 +8

c= -2

Thus, the equation of the perpendicular bisector is y= -x -2.

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3 years ago
The scatterplot below shows the relationship between the average life of carpet
ale4655 [162]

Answer:

45665

Step-by-step explanation:

i dont think that is right

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=32.5%20%3D%2030%20%2B%20%20%5Cfrac%7B20%20-%20%2840%20%2B%20x%29%7D%7B12%7D%20%20%5Ctimes%2010
erastova [34]

Answer:

x =  - 23

Step-by-step explanation:

\frac{20 - (40 + x)}{12}  \times 10 = 32.5 - 30 = 2.5

=  >  \frac{20 - (40 + x)}{12}  = 0.25

=  >  \frac{20 - (40 + x)}{12}  =  \frac{1}{4}

=  > 20 - (40 + x) =  \frac{12}{4}  = 3

=  > 40 + x = 20 - 3 = 17

=  > x = 17 - 40 =  - 23

7 0
3 years ago
6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP
adelina 88 [10]

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = 360^{o} - 324^{o}

     = 036^{o}

Sum of angles in a triangle = 180^{o}

<P + <Q + <S = 180^{o}

048° + 036^{o} + <S = 180^{o}

84^{o} + <S = 180^{o}

<S  = 180^{o} -  84^{o}

    = 96^{o}

<S = 96^{o}

Applying the Sine rule,

\frac{y}{Sin P} = \frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{y}{Sin P} = \frac{z}{Sin S}

\frac{y}{Sin 48^{o} } = \frac{17}{Sin 96^{o} }

\frac{y}{0.74314} = \frac{17}{0.99452}

⇒ y = \frac{12.63338}{0.99452}

       = 12.703

y = 12.70 km

\frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{x}{Sin 36^{o} } = \frac{17}{Sin 96^{o} }

\frac{x}{0.58779} = \frac{17}{0.99452}

⇒ x = \frac{9.992430}{0.99452}

      = 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

6 0
3 years ago
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