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Natasha2012 [34]
3 years ago
8

If f is translated 4 units down to create g, which function represents g? A. g(x) = f(x) - 4 B. g(x) = f(x - 4) C. g(x) = f(x +

4) D. g(x) = f(x) + 4 v

Mathematics
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

g(x)=f(x)-4

Step-by-step explanation:

we are given

f is translated 4 units down to create g

we know that

whenever we translate any function by 'a' units down , we always subtract that function by 'a' units

Suppose, f(x) is translated down by 'a' units

so, we get

=f(x)-a

now, we have

f is translated 4 units down

so, we get

=f(x)-4

and it is creating g(x)

g(x)=f(x)-4


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Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of
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Answer:

The "probability that a given score is less than negative 0.84" is  \\ P(z.

Step-by-step explanation:

From the question, we have:

  • The random variable is <em>normally distributed</em> according to a <em>standard normal distribution</em>, that is, a normal distribution with \\ \mu = 0 and \\ \sigma = 1.
  • We are provided with a <em>z-score</em> of -0.84 or \\ z = -0.84.

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A z-score is a standardized value, i.e., one that we can obtain using the next formula:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

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  • And we already know that \\ \mu and \\ \sigma are the mean and the standard deviation, respectively, of the <em>normal distribution</em>.

A <em>z-score</em> represents the <em>distance</em> from \\ \mu in <em>standard deviations</em> units. When the value for z is <em>negative</em>, it "tells us" that the raw score is <em>below</em> \\ \mu. Conversely, when the z-score is <em>positive</em>, the standardized raw score, <em>x</em>, is <em>above</em> the mean, \\ \mu.

Solving the question

We already know that \\ z = -0.84 or that the standardized value for a raw score, <em>x</em>, is <em>below</em> \\ \mu in <em>0.84 standard deviations</em>.

The values for probabilities of the <em>standard normal distribution</em> are tabulated in the <em>standard normal table, </em>which is available in Statistics books or on the Internet and is generally in <em>cumulative probabilities</em> from <em>negative infinity</em>, - \\ \infty, to the z-score of interest.

Well, to solve the question, we need to consult the <em>standard normal table </em>for \\ z = -0.84. For this:

  • Find the <em>cumulative standard normal table.</em>
  • In the first column of the table, use -0.8 as an entry.
  • Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
  • The intersection of these two numbers "gives us" the cumulative probability for z or \\ P(z.

Therefore, we obtain \\ P(z for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".

This represent the area under the <em>standard normal distribution</em>, \\ N(0,1), at the <em>left</em> of <em>z = -0.84</em>.

To "draw a sketch of the region", we need to draw a normal distribution <em>(symmetrical bell-shaped distribution)</em>, with mean that equals 0 at the middle of the distribution, \\ \mu = 0, and a standard deviation that equals 1, \\ \sigma = 1.

Then, divide the abscissas axis (horizontal axis) into <em>equal parts</em> of <em>one standard deviation</em> from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).  

Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, \\ -\sigma, from it). All the area to the left of this value must be shaded because it represents \\ P(z and that is it.

The below graph shows the shaded area (in blue) for \\ P(z for \\ N(0,1).

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Answer:

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Step-by-step explanation:

We know that the area of a circle is $pi*r^2$, and we also know that the diameter of a circle is equal to $2r$.

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