All categories

3 years ago

Which line best represents the line of best fit for this scatter plot?

Mathematics

1 answer:

Lerok [7]3 years ago

7 0

Answer: line R

Step-by-step explanation: because it crossing through the starting and ending points

You might be interested in

Expand and simplify 3(5x + 2) +2(x-1)

RoseWind [281]

Answer:

17x+4

Step-by-step explanation:

Just Multiply the Expression:

15x+6+2x-2

17x+4

8 0

4 years ago

Read 2 more answers

ted says that 3/10 multiplied by 100 equals 300000 is he Correct you the place value chart to explain your answer

kolezko [41]

Ted is incorrect because 3/10 is the same as .30 which is then multiplied by 100 and given the product of 30.
.3×100=30

3 0

3 years ago

Read 2 more answers

I have a list of expressions to simplify. pls help!

Nezavi [6.7K]

Answer:

1. 80

2. -13

3. -11

4. 21

5. -18

6. -8

7. -6

8. 2

9. -23

10. 10

11. -36

12. 28

13. 12

14. 9

15. -91

16. -10

17. 12

18. -9

19. 5

20. 44

21. 0.7273

22. -169

23. 15

24. -48

25. -7

26. 2

27. 11

28. -9

29. 104

30. -1

Step-by-step explanation:

Plz mark brainliest thnxxx

7 0

3 years ago

Pls answer this asap

Serga [27]

Answer:

its c or d

Step-by-step explanation:

6 0

3 years ago

Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.

solong [7]

Answer:

$A)\,\,det(A)=1$

$B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]$

Step-by-step explanation:

$det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|$

Expanding with first row

$det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1$

To find inverse we first find cofactor matrix

$C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\$

$C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\$

Cofactor matrix is

$C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]$

4 0

3 years ago