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3 years ago

1. Describe the transformations necessary to transform the graph of

Mathematics

1 answer:

lana66690 [7]3 years ago

6 0

Answer:

- The function $f(x)$ must be shifted right 1 unit.

- The function $f(x)$ must be shifted up 4 units.

- The function $f(x)$ must be stretched vertically by a factor of 3.

Step-by-step explanation:

There are some transformations for a function f(x):

If $f(x)+k$ , then the function is shifted up "k" units.

If $f(x)-k$ , then the function is shifted down "k" units.

If $f(x+k)$ , then the function is shifted left "k" units.

If $f(x-k)$ , then the function is shifted right "k" units.

If $bf(x)$ and $b>1$ , then the function is stretched vertically by a factor of "b".

If $bf(x)$ and $0$ , then the function is compressed vertically by a factor of "b".

In this case we have the following parent function $f(x)$ :

$f(x)=x^2$

And the function $g(x)$ :

$g(x) = 3(x - 1)^2 + 4$

Based on the explained before, we can describe the transformations necessary to transform the graph of $f(x)$ to the graph of $g(x)$ :

- The function $f(x)$ must be shifted right 1 unit.

- The function $f(x)$ must be shifted up 4 units.

- The function $f(x)$ must be stretched vertically by a factor of 3.

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For the answer to the question above,
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3 years ago

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The movie theater has 250 seats 225 seats were sold for the carriage showing what percent of seats are empty?

Nezavi [6.7K]

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3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

What is the ratio of 3/20

Sveta_85 [38]

The ratio of 3/20 is 3:20

5 0

3 years ago

A bank's loan officer rates applicants for credit. the ratings are normally distributed with a mean of 200 and a standard deviat

VMariaS [17]

The probability is 0.4332.

We want to find the z-scores associated with the ends of the ranges. The z-score for 200 would be 0, since it is 0 away from the mean. To find the z-score of 275 we use the formula:

$z=\frac{X-\mu}{\sigma}
\\
\\z=\frac{275-200}{50}=\frac{75}{50}=1.5$

Using a z-table (http://www.z-table.com) we can find the area under the curve to the left of 1.5 (everything less than that, so every score less than 275). This is 0.9338. However, we only want the range from the mean, 200, up to this point. This means we subtract the area to the left of 200, 0.5 (half of the data goes from the mean left), from this:

0.9338-0.5=0.4338.

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4 years ago