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3 years ago

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Given that the first step of the sequence of transformations that maps ABCD to PQRST is a reflection, identify the the transform

ation in coordinate notation.

1 answer:

olga55 [171]3 years ago

4 0

Answer:

d

Step-by-step explanation:

You might be interested in

The end points of one side of a regular pentagon are (-1,4) and (2,3). What is the perimeter of the pentagon?

marissa [1.9K]

To determine the perimeter of the pentagon, you must first calculate a side length of it. Let's name the coordinates A(-1,4) and B(2,3).

To figure out how far the points are from each other, you have to use the distance formula:
$D_{AB} = \sqrt{( x_{2}-x_{1})^2+{(y_{2}-y_{1})^2}
$
$x_{1} =1, x_{2} =-2, y_{1} =2, y_{2} =3$
$D_{AB}= \sqrt{(2--1)^2+{(3-4)^2}$
D_{AB}= \sqrt{(2--1)^2+{(3-4)^2}
$D_{AB}= \sqrt{(2+1)^2+{(3-4)^2}$
$D_{AB}= \sqrt{(3)^2+(-1)^2}$
$D_{AB}= \sqrt{9+1}$
$D_{AB}= \sqrt{10}$

Now, the formula for the perimeter of a pentagon is
P = 5×side length

So...
Perimeter = 5× $\sqrt{10}$

The answer is (2)

8 0

3 years ago

A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The s

Fofino [41]

Answer:

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question, we have that:

$\mu = 270, \sigma = 90, n = 18, s = \frac{90}{\sqrt{18}} = 21.2$

What is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days?

This is the pvalue of Z when $X = 260$ . So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{260 - 270}{21.2}$

$Z = -0.47$

$Z = -0.47$ has a pvalue of 0.3192.

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

5 0

3 years ago

Help me with this and I need an explanation I’m confused.

xxTIMURxx [149]

7x10x and use yout braing

5 0

3 years ago

Help me out plssssssss

max2010maxim [7]

The answer to this is D

8 0

3 years ago

When consumers apply for credit, their credit is rated using FICO (Fair, Isaac, and Company) scores. Credit ratings are given be

Flura [38]

Answer:

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is $\bar X= 660.313$

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . The degrees of freedom are given by:

$df=n-1=16-1=15$

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for $t_{\alpha/2}=-2.95$

"=T.INV(1-0.005,15)" for $t_{1-\alpha/2}=2.95$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we find the limits we got:

$660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588$

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.

4 0

3 years ago