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faust18 [17]
2 years ago
11

Which sequence of transformations produces an image that is not congruent to the original figure? A. A translation of 4 units to

the right followed by a dilation of 2. B. A rotation of 90 degrees clockwise followed by a translation of 2 units to the left. . C. A translation of 3 units to the left followed by a reflection across the y-axis. D. A reflection across the y-axis followed by a rotation of 90 degrees counterclockwise.
Mathematics
2 answers:
Nataliya [291]2 years ago
4 0

The <em>correct answer</em> is:


A) A translation of 4 units to the right followed by a dilation of 2.


Explanation:


3 of our transformations are called isometries. These are transformations that preserve the size and shape of the original figure; they just change the position or orientation of it. Translations, reflections and rotations are all isometries. We know that isometries preserve congruence, since they maintain the size and shape of the original figure.


A dilation is not an isometry. This is because a dilation changes the size of a figure. Since the new figure will not be the same size as the original, they will not be congruent.

777dan777 [17]2 years ago
4 0

Answer:

Correct option is A. A translation of 4 units to the right followed by a dilation of 2.

Step-by-step explanation:

In options B, C and D, rotation, translation and reflection takes place. Therefore, they are all congruent to the original figure.

In option A, translation of 4 units to the right followed by a dilation of 2 takes place. So, it changes the size of the original figure and hence it is not congruent to the original figure.

Hence, the correct option is A.

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Answer:

Step-by-step explanation:

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An online retail company tracked total number of purchases monthly. In June 2007 they were 4,500 purchases and in September 2008
Verizon [17]

Complete Question:

An online retail company tracked total number of purchases monthly. In June 2007 they were 4,500 purchases and in September 2008 there were 5,100. The purchases followed a linear model of the form p = S ( t ) = r t + i where p purchases is in 1000's and t is months where January 2007=0

a) What is the constant rate of sales

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Answer:

a) Rate of sales, r = 40

b) Linear model function, p = s(t) = 40t + 2100

Step-by-step explanation:

p = S ( t ) = r t + i ......................(1)

Since January 2007 corresponds to t = 0, June 2007 corresponds to t = 6

Purchase, p = S(6) = 4500

Substituting these values into equation (1):

6r + i = 4500......................(2)

September 2008 corresponds to t = 21

Purchase, p = S(21) = 5100

21r + i = 5100......................(3)

Subtracting equation (2) from equation (3):

15r = 600

r = 600/15

r = 40

a) Therefore, rate of sale, r = 40

Substituting the value of r into equation (2)

6(40) + i = 4500

i = 4500 - 2400

i = 2100

b) The linear model function is:

p = s(t) = 40t + 2100

6 0
1 year ago
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choli [55]

Answer:

We conclude that the mean waiting time is more than or equal to 5 minutes at 5% level of significance.

Step-by-step explanation:

We are given that waiting time is defined as the time the customer enters the line until he or she reaches the teller window.  

A random sample of 15 customers is selected. Results found that the sample mean waiting time was 4.287 minutes with a sample standard deviation of 1.638 minutes.

Let \mu = <u><em>mean waiting time.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 5 minutes     {means that the mean waiting time is more than or equal to 5 minutes}

Alternate Hypothesis, H_A : \mu < 5 minutes      {means that the mean waiting time is less than 5 minutes}

The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;

                              T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean waiting time = 4.287 minutes

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            n = sample of customers = 15

So, <u><em>the test statistics</em></u>  =  \frac{4.287 -5}{\frac{1.638}{\sqrt{15} } }  ~ t_1_4

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The value of t test statistics is -1.686.

Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.686 > -1.761, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean waiting time is more than or equal to 5 minutes.

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