Answer:
A. congruent -SSS
Step-by-step explanation:
Picture below
The sum of three consecutive numbers is -9 what are the numbers
2 answers:
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A) 2x -3(3x -7) = 7 . . . . substitute for y
.. -7x = -14 . . . . . . . . . . . subtract 21
.. x = 2 . . . . . . . . . . . . . divide by -7
.. y = 3*2 -7 = -1 . . . . . . substitute for x
(x, y) = (2, -1)
B) 5(5x -2y) -2(3x -5y) = 5(8) -2(1) . . . . subtract 2 times 2nd equation from 5 times 1st
.. 19x = 38 . . . . simplify
.. x = 2 . . . . . . . divide by 19
.. y = (5x -8)/2 = 1 . . . . solve 1st equation for y, substitute x
(x, y) = (2, 1)
C) See the first graph.
(x, y) = (-4, 1)
D) See the second graph.
(x, y) = (3, 2)
.. -7x = -14 . . . . . . . . . . . subtract 21
.. x = 2 . . . . . . . . . . . . . divide by -7
.. y = 3*2 -7 = -1 . . . . . . substitute for x
(x, y) = (2, -1)
B) 5(5x -2y) -2(3x -5y) = 5(8) -2(1) . . . . subtract 2 times 2nd equation from 5 times 1st
.. 19x = 38 . . . . simplify
.. x = 2 . . . . . . . divide by 19
.. y = (5x -8)/2 = 1 . . . . solve 1st equation for y, substitute x
(x, y) = (2, 1)
C) See the first graph.
(x, y) = (-4, 1)
D) See the second graph.
(x, y) = (3, 2)
Hi!
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.