y-intercept: Let x = 0 and solve for y:
(x-1)^2 + (y-2)^2 = 10 => (-1)^2 + y^2 - 4y + 4) = 10
=> 1 + y^2 - 4y + 4 = 10, or y^2 - 4y -5 = 0
The solutions of this quadratic are y = 5 and y = -1.
Thus, the y-intercepts are (0, 5) and (0, -1).
Now find the x-intercepts: Let y = 0 and solve the resulting equation for x:
(x-1)^2 = 10 - (-2)^2, or (x-1)^2 = 10 - 4 = 6.
Taking the sqrt of both sides, x - 1 = plus or minus sqrt(6), or:
x = 1 +√6 and x = 1 - √6, so that the x-intercepts
are (1+√6, 0) and (1-√6, 0).
Unknown answer, not enough info.
Answer:
J(-2, -1), K(4, -5), L(0, -5)
The midpoint of segment JL is
(-2 + 0)/2, (-1 + (-5))/2) = (-2/2, -6/2) = (-1, -3)
The midpoint is
(-2 + 4)/2, (-1 + (-5))/2) = (2/2, -6/2) = (1, -3)
A: (-1, -3), (1, -3)
Step-by-step explanation:
Answer:
Hence, data does not suggest the true average gap detection threshold for CTS subjects exceeds that for normal subjects.
Step-by-step explanation:
H0 : μ1 = μ2
H1 : μ1 < μ2
Given :
m = 8 ; x1 = 1.71 ; s1 = 0.53
n = 10 ; x2 = 2.53 ; s2 = 0.87
The test statistic :
(x1 - x2) / √(s1²/m + s2²/n)
(1.71 - 2.53) / √(0.53²/8 + 0.87²/10)
-0.82 / √0.1108025
Test statistic = - 0.82 / 0.3328700
Test statistic = - 2.463
The degree of freedom using the conservative approach :
Smaller of (10 - 1) or (8 - 1)
df = 7
TCritical value(0.01, 7) = 2.998
Decision region :
Reject H0 if |Test statistic| > |critical value|
Since, 2.463 < 2.998 ; WE fail to reject H0 ; Hence result is not significant at α = 0.01
62 multiplied by 3 is 186
Add on 31 for the half hour
and you get 217
