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1 year ago

Which is the best linear model for the data shown in the scatter plot?

Mathematics

1 answer:

Studentka2010 [4]1 year ago

3 0

The best linear model for the data that's illustrated will be y = 10x + 5

<h3>What is a linear model?</h3>

It should be noted that a linear model simply means model where the terms are added. It should be noted that this can be used in connection with the regression model.

Here, the best linear model for the data will be:

= (10 × x) + 5

= 10x + 5

Therefore, the best linear model for the data that's illustrated will be y = 10x + 5.

Learn more about model on:

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Given: AABC, AC = 5<br> m C = 90°<br> m A= 22°<br> Find: Perimeter of AABC<br> A<br> C<br> B

Leto [7]

9514 1404 393

Answer:

perimeter ≈ 12.4 units

Step-by-step explanation:

The side adjacent to the angle is given. The relationships useful for the other two sides are ...

Tan = Opposite/Adjacent

Cos = Adjacent/Hypotenuse

From these, we have ...

opposite = 5·tan(22°) ≈ 2.02

hypotenuse = 5/cos(22°) ≈ 5.39

Then the perimeter is ...

P = a + b + c = 2.02 + 5 + 5.39 = 12.41

The perimeter of ∆ABC is about 12.4 units.

7 0

2 years ago

The sum of two numbers is 39 and the difference is 3. What are the numbers?

frosja888 [35]

Answer: 21 & 18

Step-by-step explanation:

Sum of two numbers is 39

Difference is 3

Let the numbers be x & y

X+y=39........equation 1

X-y=3...........equation 2

X=3+y............equation 3

Substitute equation 3 into equation 1

(3+y)+y=39

3+y+y=39

3+2y=39

2y=39-3

2y=36

Y=36/2

Y=18

Substitute for y in equation 2

X-y=3

X-18=3

X=3+18

X=21

The two numbers are 21 & 18

8 0

3 years ago

Does anybody know how to solve this

kirill [66]

Answer:

Question 11.1

At 7 : 00 pm in Sydney, it will be 10 : 00 pm in Berlin.

Question 11.2

Place | Time

Sydney | 5 : 00 pm

Berlin | 8 : 00 pm

It would be a good time for Mark and Hans to cha,t when it's 5 : 00 pm in Sydney and 8 : 00 pm in Berlin.

hope that helps ...

5 0

2 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago

If my equation is ax^2+bx+c=0 how would the c end up on the right side and become -c?

Vaselesa [24]

By subtracting c on both sides

7 0

3 years ago

Which is the best linear model for the data shown in the scatter plot?​

Which is the best linear model for the data shown in the scatter plot?