Answer: The set does not have a solution
Step-by-step explanation:
Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5
Putting this value of x in eq. 2 we get
-2y + 2z = -1-(7/5) or
2y - 2z = 12/5 or 5y - 5z = 6
Multiplying eq. 1 by 2 we get
4x + 2y - 2z = 6
adding this with eq. 2 we get 5x = 5 or x = 1
As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.
Same can be verified by using different sets of two equations also.
Answer:
B
Step-by-step explanation:
Data set D does not contain the value 128, which is the median value.
Data set C does not contain the outlier value 91.
Data set A contains value 168, which does not show up on the plot.
The only remaining choice is B.
_____
In order, the data values of set B are ...
... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152
The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)
Answer:
x=18
Step-by-step explanation:
set up the equation:
DBC+DBC=ABC
79+79=(9x-4)
Simplify both sides of the equation.
79+79=9x−4
79+79=9x+−4
(79+79)=9x−4(Combine Like Terms)
158=9x−4
158=9x−4
Flip the equation.
9x−4=158
Add 4 to both sides.
9x−4+4=158+4
9x=162
Divide both sides by 9.
9x
/9 = 162
/9
x=18
