The critical value for a hypothesis test _______.
1 answer:
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The curve has been attached and the answer choices are:
y = 3x² – 2x + 1
y = 3x² – 6x + 3
y = 3x²<span> – 7x + 1
</span>
The attached graph has a vertex in the first quadrant. Therefore, the coordinates of the vertex would be both positive.
Let's start with first equation:
y = 3x² – 2x + 1
using the equation of axis:
x = -b/2a
x = 2/6
x = 1/3
SUbstituting the value of x in the main equation to get the y-coordinate of the vertex.
y = 3(1/3)² – 2(1/3) + 1
y = 3/9 – 2/3 + 1
y = 1/3 – 2/3 + 1
y = (1 - 2 + 3)/3
y = 2/3
Hence, the vertex would be:
(h,k) = (1/3 , 2/3)
Also, the leading coefficient is positive, so the parabola would be concave up.
Thus the final answer choice will be:
y = 3x² – 2x + 1
y = 3x² – 2x + 1
y = 3x² – 6x + 3
y = 3x²<span> – 7x + 1
</span>
The attached graph has a vertex in the first quadrant. Therefore, the coordinates of the vertex would be both positive.
Let's start with first equation:
y = 3x² – 2x + 1
using the equation of axis:
x = -b/2a
x = 2/6
x = 1/3
SUbstituting the value of x in the main equation to get the y-coordinate of the vertex.
y = 3(1/3)² – 2(1/3) + 1
y = 3/9 – 2/3 + 1
y = 1/3 – 2/3 + 1
y = (1 - 2 + 3)/3
y = 2/3
Hence, the vertex would be:
(h,k) = (1/3 , 2/3)
Also, the leading coefficient is positive, so the parabola would be concave up.
Thus the final answer choice will be:
y = 3x² – 2x + 1