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otez555 [7]
3 years ago
15

Suppose f(x)=x^2. what is the graph of g(x)=1/3 f(x)

Mathematics
1 answer:
saw5 [17]3 years ago
6 0

Step-by-step explanation:

f(x) = x^2

g(x)= 1/3 f(x) = 1/3 x^2

First lets make a table for f(x)= x^2

Lets pick some numbers for x  and find out y

x           y=x^2

-3           (-3)^2 = 9

-1              1

0             0

1              1

3             9

Now we use the same x  values and make a table for g(x)

x         g(x)= \frac{1}{3} x^2

-3          g(x)= \frac{1}{3}(-3)^2=3

-1          g(x)= \frac{1}{3}(-1)^2=\frac{1}{3}

0             0

1           g(x)= \frac{1}{3}(1)^2=\frac{1}{3}

3          g(x)= \frac{1}{3}(3)^2=3

Graph the above table



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There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters tha
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Answer:

0.6826 = 68.26% probability that the first machine produces an acceptable cork.

0.933 = 93.3% probability that the second machine produces an acceptable cork.

The second machine is more likely to produce an acceptable cork.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

What is the probability that the first machine produces an acceptable cork?

The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.10 cm, which means that \mu = 3, \sigma = 0.1

Acceptable between 2.9 and 3.1, which means that this probability is the pvalue of Z when X = 3.1 subtracted by the pvalue of Z when X = 2.9.

X = 3.1

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.1 - 3}{0.1}

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X = 2.9

Z = \frac{X - \mu}{\sigma}

Z = \frac{2.9 - 3}{0.1}

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Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

0.6826 = 68.26% probability that the first machine produces an acceptable cork.

What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.)

For the second machine, we have that \mu = 3.04, \sigma = 0.04. Same probability we have to find out. So

X = 3.1

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.1 - 3.04}{0.04}

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Z = \frac{X - \mu}{\sigma}

Z = \frac{2.9 - 3.04}{0.04}

Z = -3.5

Z = -3.5 has a pvalue of 0.0002

0.9332 - 0.0002 = 0.933

0.933 = 93.3% probability that the second machine produces an acceptable cork.

Which machine is more likely to produce an acceptable cork?

Second one has a higer probability, so it is more likely to produce an acceptable cork.

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