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2 years ago

Is there a series of rigid transformations that could map A

Mathematics

2 answers:

topjm [15]2 years ago

7 0

Answer:

your answer is C. Yes, Triangle KLP can be reflected across the line containing KP and then translated so that P is mapped to M on edg2020

MaRussiya [10]2 years ago

3 0

Answer:

Yes, A KLP can be reflected across the line containing KP and then translated so that Pis mapped to M.

Step-by-step explanation:

The figure shows two congruent by HA theorem (they have congruent hypotenuses and a pair of congruent angles adjacent to the hypotenuses) triangles KLP and QNM.

A rigid transformation is a transformation which preserves lengths. Reflection, rotation and translation are rigit transformations.

If you reflect triangle KLP across the leg KP and translate it up so that point P coincides with point M , then the image of triangle KLP after these transformations will be triangle QNM.

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Match the following items by evaluating the expression for x = -6.

tigry1 [53]

Answer:

What are the following items?

Step-by-step explanation:

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2 years ago

Linear Pairs and Vertical Angles

goblinko [34]

Answer:

angles 1 and 3 are 29 degrees

angles 2 and 4 are 151 degrees

Step-by-step explanation:

make angles 1 and 3 equal echo there and solve for x

use x to find the angle

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2 years ago

Write the perimeter of the floor plan shown as an algebraic expression in x.

Nimfa-mama [501]

The perimeter is the sum of the enclosing side.

From the figure, the perimeter is
P = 11 + (x-2) + (11-3) + [(x-2) - (x-11)] + (x-11)
= 11 + x - 2 + 8 + 9 + x - 11
= 2x + 15

Answer: 2x + 15

5 0

2 years ago

A professor, transferred from Toronto to New York, needs to sell his house in Toronto quickly. Someone has offered to buy his ho

I am Lyosha [343]

Answer:

(a) = 40%

(b) = 28%

Standard deviation = $7,216.88

Step-by-step explanation:

This is a normal distribution with a = 210,000 and b =235,000

(a) The probability that he will get at least $225,000 for the house is:

$P(X\geq 225,000) =1 -\frac{225,000-a}{b-a} =1-\frac{225,000-210,000}{235,000-210,000} \\P(X\geq 225,000) =0.4= 40\%$

(b)The probability he will get less than $217,000 is:

$P(X\leq 217,000) =\frac{217,000-a}{b-a} =\frac{217,000-210,000}{235,000-210,000} \\P(X\leq 217,000) =0.28= 28\%$

(c) The expected value (E) and the standard deviation (S) are:

$E=\frac{a+b}{2}=\frac{210,000+235,000}{2}\\ E=\$222,500\\S=\frac{b-a}{\sqrt{12}}=\frac{235,000-210,000}{\sqrt{12}}\\S=\$7,216.88$

4 0

2 years ago

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

arlik [135]

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

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".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

Now we have everything in order to replace into formula (1):

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$

$-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087$

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

4 0

1 year ago