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SpyIntel [72]
1 year ago
8

Which statement is true about a trend line? A regression line and a trend line are opposite terms. A regression line and trend l

ine are equivalent terms. A trend line represents only the smallest data points. A trend line represents only the largest data points.
Mathematics
2 answers:
Naily [24]1 year ago
8 0

Answer:

b

Step-by-step explanation:

kicyunya [14]1 year ago
5 0

Answer:

A is incorrect!  <u>B. should be the correct answer</u>

Step-by-step explanation

After some research, i found that the correct answer is most likely <u><em>B.  a regression line and trend line are equivalent terms</em></u>

<em />

<em>A trendline and a regression can be the same. </em>

<em> A regression line is based upon the best fitting curve Y= a + bX Most often it’s a least-squares fit (where the squared distances from the points to the line (along the Y-axis) is minimized). </em>

<em> It can be quadratic or logistic or otherwise, but most often it is linear. </em>

<em> A trendline is often constructed by smoothing of the results, making it less peaked. (often by using a moving average); but can also come from ARIMA projections or curve fitting techniques (such as regression).</em>

<em />

Let me know if i helped you!

<em />

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A log is 16m long, correct to the nearest metre. It has to be cut into fence posts which must be 70cm long, correct to the neare
dmitriy555 [2]

Answer:

25 posts

Step-by-step explanation:

So the number of fence post would be the total length of the log divided by the length of each post. As the log is 16m and is corrected to the nearest metre, it could possibly be 16.499m. As for the post that is 70 cm long and corrected to the nearest 10cm, it may as well be 65 cm (or 0.65m) each post

So the max number of fence point once can possibly cut from the log would be

16.499 / 0.65 = 25 posts

5 0
2 years ago
Plz answer quickly!!!!!!!
ValentinkaMS [17]

Answer:

<em>(A). x ≥ </em>\frac{16}{5}<em> and x ≤ - </em>\frac{3}{4}<em> </em>

Step-by-step explanation:

5x - 4 ≥ 12 ⇔ 5x ≥ 16 ⇒ <em>x ≥ </em>\frac{16}{5}<em> </em>

<em>and </em>

12x + 5 ≤ - 4 ⇔ 12x ≤ - 9 ⇔ 4x ≤ - 3 ⇒ <em>x ≤ - </em>\frac{3}{4}

6 0
2 years ago
Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
2 years ago
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Using the power series methods solve the 1st order Lane-Emden Equation:
nikklg [1K]

Answer:

Step-by-step explanation:

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Differentiating equation (1) n times by Leibnitz theorem, gives:

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2y(n) = -ny(n - 1)

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For n = 1: y = 0

For n = 2: y(1) = -y

For n = 3: -3y(2)/2

For n = 4: -2y(3)

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