Solve this please!!!
1 answer:
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Answer:
The answer is parabola; (y')² - 8x' = 0 ⇒ answer (b)
Step-by-step explanation:
* At first lets talk about the general form of the conic equation
- Ax² + Bxy + Cy² + Dx + Ey + F = 0
∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.
∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.
∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.
* Now we will study our equation:
* x² - 8y = 0
∵ A = 1 , B = 0 , C =
∴ B² - 4AC = (0)² - 4(1)(0) = 0
∵ B² - 4AC = 0
∴ it will be a parabola.
∵ Ф = 90°
* The point (x , Y) will be (x' , y')
∵ x = x'cosФ - y'sinФ and y = x'sinФ + y'cosФ
∵ cos(90) = 0 and sin(90) = 1
∴ x = -y' and y = x'
* lets substitute x and y in the first equation
∴ (-y')² - 8(x') = 0
∴ (y')² - 8x' = 0
* We notice that the x' took the place of y and y' took the place of x
∴ The parabola rotated around the origin by 90°
∴ The equation of the parabola is (y')² - 8x' = 0
* The answer is parabola, with angle of rotation 90°
* The equation is (y')² - 8x' = 0
* Look to the graph
- The blue is x² - 8y = 0
- The green is (y')² - 8x' = 0
Answer:
a)
i. x=- - - - - -
ii. o=- - - - - -
iii. Sx=- - - - -
b) The random variable X signifies the weights of the bags of candies which are selected at random.
c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.
d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.
N(X, o/)
So, the distribution is
N(2, 0.12/)
e)
i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411
iii. The error bound is 0.0411
f)
i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582
iii. The error bound is 0.0582
g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.
Hence the interval in part is larger than the one in part (e).
h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.
Step-by-step explanation:
a)
i. x=- - - - - -
ii. o=- - - - - -
iii. Sx=- - - - -
b) The random variable X signifies the weights of the bags of candies which are selected at random.
c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.
d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.
N(X, o/)
So, the distribution is
N(2, 0.12/)
e)
i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411
iii. The error bound is 0.0411
f)
i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582
iii. The error bound is 0.0582
g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.
Hence the interval in part is larger than the one in part (e).
h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.