Answer:
hmmm 38
Step-by-step explanation:
?
Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
D.
1/5 of 65 is
65 × 1 / 5
65 / 5
= 13
65 and 13 add up to 78.
65 is the highest number out of both numbers.
♧
I assume the first equation is supposed to be
and not
As an augmented matrix, this system is given by
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C4%26-2%264%2612%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/2:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C2%26-1%262%266%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%265%265%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/5:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 3) to row 1, and add 3(row 3) to row 2:
![\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%260%2611%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -3(row 2) to row 1:
![\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D-1%260%260%26-1%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 1 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%26-1%260%262%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multipy through row 2 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%261%260%26-2%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
The solution to the system is then
Answer:
the first cell next to 3 (time hours x) is 300
the point is (3, 300)
the second cell is 1000
the point is (10,1000)
the values for the second table should be the same if they have the same unit rate which is 100 watts per hour
Step-by-step explanation:
If there is 100 watts perhour just multiply the number of hours by this unit
