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irinina [24]
2 years ago
15

Application of the least squares method results in values of the y-intercept and the slope that minimizes the sum of the squared

deviations between the _____.
a. actual values of the independent variable and the predicted values of the dependent variable.
b. observed values of the dependent variable and the predicted values of the dependent variable
c. observed values of the independent variable and the predicted values of the independent variable
d. None of the answers is correct
Mathematics
1 answer:
vaieri [72.5K]2 years ago
5 0

Answer:

c. observed values of the independent variable and the predicted values of the independent variable

Step-by-step explanation:

This helps us, for example, find the values of y in a y = f(x) equation. y is dependent of x. So x is the independent variable and y the dependent. Obviously, this system is used for way more complex equations, in which is hard to find an actual pattern for y, so we use this method to compare the predicted values of y to the observed.

The correct answer is:

c. observed values of the independent variable and the predicted values of the independent variable

You might be interested in
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section
LUCKY_DIMON [66]

Answer:

y2 = C1xe^(4x)

Step-by-step explanation:

Given that y1 = e^(4x) is a solution to the differential equation

y'' - 8y' + 16y = 0

We want to find the second solution y2 of the equation using the method of reduction of order.

Let

y2 = uy1

Because y2 is a solution to the differential equation, it satisfies

y2'' - 8y2' + 16y2 = 0

y2 = ue^(4x)

y2' = u'e^(4x) + 4ue^(4x)

y2'' = u''e^(4x) + 4u'e^(4x) + 4u'e^(4x) + 16ue^(4x)

= u''e^(4x) + 8u'e^(4x) + 16ue^(4x)

Using these,

y2'' - 8y2' + 16y2 =

[u''e^(4x) + 8u'e^(4x) + 16ue^(4x)] - 8[u'e^(4x) + 4ue^(4x)] + 16ue^(4x) = 0

u''e^(4x) = 0

Let w = u', then w' = u''

w'e^(4x) = 0

w' = 0

Integrating this, we have

w = C1

But w = u'

u' = C1

Integrating again, we have

u = C1x

But y2 = ue^(4x)

y2 = C1xe^(4x)

And this is the second solution

5 0
2 years ago
I increase a number by 24% the answer is 6014 what number did I start with
miskamm [114]

Answer:

1443.36

Step-by-step explanation:

6014*.24=1443.36

Hope it helps

8 0
2 years ago
A boat tour guide expects his tour to travel at a rate of x mph on the first leg of the trip. On the return route, the boat trav
Georgia [21]

<em><u>The intervals included in solution are:</u></em>

\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0

<em><u>Solution:</u></em>

Given that,

A boat tour guide expects his tour to travel at a rate of x mph on the first leg of the trip

On the return route, the boat travels against the current, decreasing the boat's rate by 10 mph

The group needs to travel an average of at least 24 mph

<em><u>Given inequality is:</u></em>

\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}

<em><u>We have to solve the inequality</u></em>

\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}\\\\\frac{2}{x}  - 10\geq \frac{2}{24}

\mathrm{Subtract\:}\frac{2}{24}\mathrm{\:from\:both\:sides}\\\\\frac{2}{x}-10-\frac{2}{24}\ge \frac{2}{24}-\frac{2}{24}\\\\Simplify\\\\\frac{2}{x}-10-\frac{2}{24}\ge \:0

\frac{2}{x}-\frac{10}{1}-\frac{2}{24} \geq 0\\\\\frac{ 2 \times 24}{x \times 24} -\frac{10 \times 24}{1 \times 24} - \frac{2 \times x }{24 \times x}\geq 0\\\\\frac{48}{24x}-\frac{240x}{24x}-\frac{2x}{24x}\geq 0\\\\Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions\\\\\frac{48-240x-2x}{24x}\geq 0\\\\Add\:similar\:elements\\\\\frac{48-242x}{24x}\ge \:0

\mathrm{Multiply\:both\:sides\:by\:}24\\\\\frac{24\left(48-242x\right)}{24x}\ge \:0\cdot \:24\\\\Simplify\\\\\frac{48-242x}{x}\ge \:0\\\\Factor\ common\ terms\\\\\frac{-2\left(121x-24\right)}{x}\ge \:0\\\\\mathrm{Multiply\:both\:sides\:by\:}-1\mathrm{\:\left(reverse\:the\:inequality\right)}

When we multiply or divide both sides by negative number, then we must flip the inequality sign

\frac{\left(-2\left(121x-24\right)\right)\left(-1\right)}{x}\le \:0\cdot \left(-1\right)\\\\\frac{2\left(121x-24\right)}{x}\le \:0\\\\\mathrm{Divide\:both\:sides\:by\:}2\\\\\frac{\frac{2\left(121x-24\right)}{x}}{2}\le \frac{0}{2}\\\\Simplify\\\\\frac{121x-24}{x}\le \:0

\mathrm{Find\:the\:signs\:of\:the\:factors\:of\:}\frac{121x-24}{x}\\

This is attached as figure below

From the attached table,

\mathrm{Identify\:the\:intervals\:that\:satisfy\:the\:required\:condition:}\:\le \:\:0\\\\0

<em><u>Therefore, solution set is given as</u></em>:

\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0

8 0
2 years ago
Read 2 more answers
Let L be the line with parametric equations x=2+t, y=1-t, z=1+3t.Let v=(1,2,0).Find vectors w1 and w2 such that v= w1 + w2, and
charle [14.2K]

Answer:

w1 = (-1/11, 1/11, -3/11)

w2 = (12/11, 21/11, 3/11)

Step-by-step explanation:

Direction ratio of w1 = Direction ratio of L (because parallel) = K+(1, -1, 3)

Let <a,b,c> be direction ratio of w2.

Then, <a,b,c>. <1,-1,3> = 0

a-b+3c = 0

v = w1 + w2

(1, 2, 0) = k(1, -1, 3) + (a, b, c)

a + k = 1

b - K = 2

c - 3k = 0

Solving 4 equations, a = 12/11, b= 21/11, c = 3/11, k=-1/11

So, w1 = -1/11(1, -1 ,3) = (-1/11, 1/11, -3/11)

w2 = (12/11, 21/11, 3/11)

6 0
2 years ago
Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).
Elan Coil [88]

Answer: \int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = \left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0

Then, replacing with point P and normal vector n:

36(x-5) + 20(y-0) + 45(z-0) = 0

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

z = \frac{180-36x-20y}{45}

When z = 0:

y = 9 + \frac{-9x}{5}

When z=0 and y=0:

x = 5

Then, triple integral is:

\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx

Calculating:

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx

\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0  {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx

\frac{1}{45} \int\limits^5_0  {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4}  } \, dx

\frac{1}{45} [30375-60750+118462.5-39150]

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx = 1087.5

<u>The volume of the tetrahedon is 1087.5 cubic units.</u>

3 0
2 years ago
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