What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0
2 answers:
Answer:
centre = (6, 1 )
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Given
x² + y² - 12x - 2y + 12 = 0
Collect the x and y terms together and subtract 12 from both sides
x² - 12x + y² - 2y = - 12
To obtain the equation in standard form use completing the square
add ( half the coefficient of the x/ y term )² to both sides
x² + 2(- 6)x + 36 + y² + 2(- 1)y + 1 = - 12 + 36 + 1
(x - 6)² + (y - 1)² = 25 ← in standard form
with centre = (6, 1 ) and radius = = 5
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Answer:
A. T > 2.539
Step-by-step explanation:
We have a hypothesis test of the mean, with unknown population standard deviation.
The hypothesis are:
From the hypothesis we can see that the test is right-tailed, so the critical value of t should be a positive value.
The degrees of freedom can be calculated as:
The significance level is 0.01, so the critical value tc should be the one that satisfies:
Looking up in a t-table, for 19 degrees of freedom, this critical value is tc=2.539.
