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Artyom0805 [142]
4 years ago
8

Which statement describes the translation of . y= -1/2(x-2) ^2 from standard position? A) Moved up and to the right. B) Moved up

and to the left. C) Moved down and to the left. D) Moved down and to the right. .

Mathematics
2 answers:
Andreas93 [3]4 years ago
5 0

Answer:  The correct option is

(D) Moved down and to the right.

Step-by-step explanation:  We are given to select the statement that best describes the translation of the following function from the standard position.

y=-\dfrac{1}{2}(x-2)^2~~~~~~~~~~~~~~~~~~~~~~(i)

This is an equation of a parabola.

The standard position of a parabola of this type is

y=x^2,

where the vertex is at (0, 0). Since the coefficient of x² is positive, so the parabola opens upwards.

Now, from equation (i), we have

the vertex of the parabola is at (2, 0) and since the coefficient of x² is negative, so the parabola opens downwards.

Also, the vertex (0, 0) has been shifted to (2, 0), so the parabola is moved 2 units to the right.

Therefore, the given parabola is moved down and 2 units to the right as compared to the standard position.

The images of the given parabola (i) and the standard parabola are shown in the attached figure below.

Thus, (d) is the correct option.

natita [175]4 years ago
4 0
The answer D) Moved down and to the right is correct.
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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
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Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

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            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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4 0
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