The "system" of equations can't be solved for (x, y), because there's no "system" of equations given. There's really only one equation.
Either of these two equations can be massaged to look exactly like the other one. And if you graph both equations, you find that they're both the same line on the graph.
ANY point on the line is a solution to both equations ... and we all know how many different points there are on a line.
Answer:
x > 8
x < -16
Step-by-step explanation:
2x + 10 > 26
2x + 10 - 10 > 26 - 10
2x > 16
2x/2 > 16/2
x > 8
2x + 10 < -26
2x + 10 - 10 < - 26 - 10
2x < -36
2x/2 < -36/2
x < -16
9514 1404 393
Answer:
- soccer: $50.31
- hockey: $57.78
Step-by-step explanation:
Let s represent the FCI for soccer. The given values tell us ...
s + (s +7.47) = 108.09
2s = 100.62 . . . . . . . . . . subtract 7.47
s = 50.31
s+7.47 = 57.78
The FCI for soccer was $50.31; the FCI for hockey was $57.78.
Adding (or subtracting) a constant to every data value adds (or subtracts) the same constant to measures of position such as center,percentiles, max or min.
Its shape and spread such as range, IQR, standard deviation remain unchanged.
When we multiply (or divide) all the data values by any constant, all measures of position (such as the mean, median, and percentiles) and measures of spread (such as the range, the IQR, and the standard deviation) are multiplied (or divided) by that same constant.
Part A:
The lowest score is a measure of location, so both addition and multiplying the lowest score of test B by 40 and adding 50 to the result will affect the lowest score of test A.
Thus, the lowest score of test A is given by 40(21) + 50 = 890
Therefore, the lowest score of test A is 890.
Part B:
The mean score is a measure of location, so both
addition and multiplying the mean score of test B by 40 and adding 50
to the result will affect the lowest score of test A.
Thus, the mean score of test A is given by 40(29) + 50 = 1,210
Therefore, the mean score of test A is 890.
Part C:
The standard deviation is a measure of spread, so multiplying the standard deviation of test B by 40 will affect the standard deviation but adding 50
to the result will not affect the standard deviation of test A.
Thus, the standard deviation of test A is given by 40(2) = 80
Therefore, the standard deviation of test A is 80.
Part D
The Q3 score is a measure of location, so both
addition and multiplying the Q3 score of test B by 40 and adding 50
to the result will affect the Q3 score of test A.
Thus, the Q3 score of test A is given by 40(28) + 50 = 1,170
Therefore, the Q3 score of test a is 1,170.
Part E:
The median score is a measure of location, so both
addition and multiplying the median score of test B by 40 and adding 50
to the result will affect the median score of test A.
Thus, the median score of test A is given by 40(26) + 50 = 1,090
Therefore, the median score of test A is 1,090.
Part F:
The IQR is a measure of spread, so multiplying the IQR of test B by 40 will affect the IQR but adding 50
to the result will not affect the IQR of test A.
Thus, the IQR of test A is given by 40(6) = 240
Therefore, the IQR of test A is 240.
