The diagram shows AMNP. Which term describes point
1 answer:
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Answer:
- (x-4.5)^2 +(y +5)^2 = 30.25
- x = (1/8)y^2 +(1/2)y +(1/2)
- y^2/36 -x^2/64 = 1
- x^2/16 +y^2/25 = 1
Step-by-step explanation:
1. Complete the square for both x and y by adding a constant equal to the square of half the linear term coefficient. Subtract 15, and rearrange to standard form.
(x^2 -9x +4.5^2) +(y^2 +10y +5^2) = 4.5^2 +5^2 -15
(x -4.5)^2 +(y +5)^2 = 30.25 . . . . . write in standard form
Important features: center = (4.5, -5); radius = 5.5.
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2. To put this in the form x=f(y), we need to add 8x, then divide by 8.
x = (1/8)y^2 +(1/2)y +(1/2)
Important features: vertex = (0, -2); focus = (2, -2); horizontal compression factor = 1/8.
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3. We want y^2/a^2 -x^2/b^2 = 1 with a=36 and b=(36/(3/4)^2) = 64:
y^2/36 -x^2/64 = 1
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4. In the form below, "a" is the semi-axis in the x-direction. Here, that is 8/2 = 4. "b" is the semi-axis in the y-direction, which is 5 in this case. We want x^2/a^2 +y^2/b^2 = 1 with a=4 and b=5.
x^2/16 +b^2/25 = 1
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The first attachment shows the circle and parabola; the second shows the hyperbola and ellipse.
Answer:
For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:
For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:
n>= 30
Step-by-step explanation:
For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:
For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:
n>= 30