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Evgesh-ka [11]
2 years ago
15

Suppose that the functions q and r are defined as follows.

Mathematics
1 answer:
satela [25.4K]2 years ago
8 0

Answer:

(r o g)(2) = 4

(q o r)(2) = 14

Step-by-step explanation:

Given

g(x) = x^2 + 5

r(x) = \sqrt{x + 7}

Solving (a): (r o q)(2)

In function:

(r o g)(x) = r(g(x))

So, first we calculate g(2)

g(x) = x^2 + 5

g(2) = 2^2 + 5

g(2) = 4 + 5

g(2) = 9

Next, we calculate r(g(2))

Substitute 9 for g(2)in r(g(2))

r(q(2)) = r(9)

This gives:

r(x) = \sqrt{x + 7}

r(9) = \sqrt{9 +7{

r(9) = \sqrt{16}{

r(9) = 4

Hence:

(r o g)(2) = 4

Solving (b): (q o r)(2)

So, first we calculate r(2)

r(x) = \sqrt{x + 7}

r(2) = \sqrt{2 + 7}

r(2) = \sqrt{9}

r(2) = 3

Next, we calculate g(r(2))

Substitute 3 for r(2)in g(r(2))

g(r(2)) = g(3)

g(x) = x^2 + 5

g(3) = 3^2 + 5

g(3) = 9 + 5

g(3) = 14

Hence:

(q o r)(2) = 14

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Answer:

\\ P(z>-2) = 0.97725 or P(x>49) is about 97.725% (or being less precise 97.5% using the <em>empirical rule</em>).

Step-by-step explanation:

We solve this question using the following information:

  1. We are dealing here with <em>normally distributed data</em>, that is "<em>the frequency distribution of the life length data is known to be mound-shaped</em>".
  2. The normal distribution is defined by two parameters: the population mean (\\ \mu) and the population standard deviation (\\ \sigma). In this case, we have that \\ \mu = 55 months, and \\ \sigma = 3 months.
  3. To find the probabilities, we have to use the <em>standard normal distribution</em>, which has \\ \mu = 0 and \\ \sigma = 1. The probabilities for this distribution are collected in the <em>standard normal table</em>, available in Statistics books or on the Internet. We can also use statistics programs to find these probabilities.
  4. For most cases, we need to use the <em>cumulative standard normal table, </em>and for this we have to previously "transform" a raw score (x) into a z-score using the next formula: \\ z = \frac{x - \mu}{\sigma} [1]. A z-score tells us the distance from the mean that a raw score is from it in <em>standard deviations units</em>. If this value is <em>negative</em>, the raw score is <em>below</em> the mean. Conversely, a <em>positive</em> value indicates that it is <em>above</em> the mean.
  5. The <em>cumulative standard normal table </em>is made for positive values of z. Since the normal distribution is <em>symmetrical</em> around the mean, we can find the negative values of z using this formula: \\ P(z [2].

Having all this information, we can solve the question.

<h3>The percentage of the manufacturer's grade A batteries that will last more than 49 months</h3>

<em>First Step: Use formula [1] to find the z-score of the raw score x = 49 months</em>.

\\ z = \frac{49 - 55}{3}

\\ z = \frac{-6}{3}

\\ z = -2

This means that the raw score is represented by a z-score of \\ z = -2, which tells us that it is<em> two standard deviations below</em> the population mean.

<em>Second Step: Consult this value in the cumulative standard normal table for z = 2 and apply the formula [2] to find the corresponding probability.</em>

For a z = 2, the probability is 0.97725.  

Then

\\ P(z

\\ P(z2)

\\ P(z2)

But we <em>are not asked</em> for P(z<-2) but for P(z>-2) = P(x>49). This probability is the <em>complement</em> of the previous result, that is

\\ P(z>-2) = 1 - P(z

\\ P(z>-2) = 1 - 0.02275

\\ P(z>-2) = 0.97725

That is, the "<em>percentage of the manufacturer's grade A batteries will last more than 49 months</em>" is

\\ P(z>-2) = 0.97725 or about 97.725%

A graph below shows this result.

Notice that if we had used the <em>68-95-99.7 rule</em> (also known as the <em>empirical rule</em>), that is, in a normal distribution, the interval between <em>one standard deviation below and above the mean</em> contains, approximately, 68% of the observations; the interval between <em>two standard deviations below and above the mean</em> contains, approximately, 95% of the observations; and the interval between <em>three standard deviations</em> below and above the mean contains, approximately, 99.7% of the observations, we could have concluded that 2.5 % of the manufacturer's grade A batteries will last <em>less</em> than 49 months, and, as a result, 1 - 0.025 = 0.975 or 97.5% will last more than 49 months.

We can conclude that with a less precise answer (but faster) because of the <em>symmetry of the normal distribution</em>, that is, 1 - 0.95 = 0.05. At both extremes we have 0.05/2 = 0.025 or 2.5% and we were asked for P(x>49) = P(z>-2) (see the graph below).

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