Which is the alternate angle of angle pqs from the given fiqure

∆ABC is mapped onto ∆A'B'C' which is then mapped onto ∆A"B"C". ∆ABC and ∆A"B"C" are shown in the diagram.

hram777 [196]

Answer:

1rotation translation dilation

2dilation

3the relationship is the shapes are the same just dilated and in different spots. The shapes are in different spots and are different sizes ut they are the same shape making them related.

Step-by-step explanation:

6 0

2 years ago

A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exa

yarga [219]

Answer:

$z= \frac{85-70}{10}=1.5$

$z= \frac{45-35}{5}=2$

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

Step-by-step explanation:

Assuming this complete question:

A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exams were scored on different scales. Remembering what he learned about the advantages of Z scores, he asks his friend for the mean and standard deviation of her class on the exam, as well as her final exam score. Here is the information:

Our student: Final exam score = 85; Class: M = 70; SD = 10.

His friend: Final exam score = 45; Class: M = 35; SD = 5.

The Z score for the student and his friend are:

A) Our student, Z= -1.07; his friend, Z= -1.14.

B) Our student, Z= 1.50; his friend, Z=2.00.

C) Our student, Z= 1.07; his friend, Z= -1.14.

D) Our student, Z= 1.07; his friend, Z= 1.50

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution

Let X the random variable that represent the scores for our student, and for this case we know that:

$E(X)= \mu =70, SD_X=\sigma=10$

The z score is given by:

$z=\frac{x-\mu}{\sigma}$

If we use this we got:

$z= \frac{85-70}{10}=1.5$

Let Y the random variable that represent the scores for his friend, and for this case we know that:

$E(Y)= \mu =35, SD_Y=\sigma=5$

The z score is given by:

$z=\frac{y-\mu}{\sigma}$

If we use this we got:

$z= \frac{45-35}{5}=2$

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

3 0

3 years ago

Lorne subtracted 6x3 – 2x + 3 from –3x3 + 5x2 + 4x – 7. Use the drop-down menus to identify the steps Lorne used to find the dif

Novay_Z [31]

Answer: Step 1: Reverse the signs of $6x^3-2x + 3$ expression.

Step 2: Removing parenthesis

Step 3: Grouping like terms

Step 4: Combing like terms

Step 5: Writing the final expression in standard form

Step-by-step explanation: First expression : $6x^3-2x + 3$

Second expression : $-3x^3+5x^2+4x-7$ .

We need to subtract $6x^3-2x + 3$ from $-3x^3+5x^2+4x-7$ .

Step 1: Reverse the signs of $6x^3-2x + 3$ expression.

$(-3x^3+5x^2+4x-7)+(-6x3+2x-3)$

Step 2: Removing parenthesis

$(-3x^3) +5x^2+4x + (-7) + (-6x^3) + 2x + (-3)$

Step 3: Grouping like terms

$[-3x^3) + (-6x^3)] + [4x + 2x] + [(-7) + (-3)] + [5x^2]$

Step 4: Combing like terms

$-9x^3 + 6x + (-10) + 5x^2$

Step 5: Writing the final expression in standard form

$-9x^3 + 5x^2 + 6x-10$

8 0

3 years ago

Read 2 more answers

Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5

Bond [772]

Answer:

The solution to the equation system given is:

<u>x = 2</u>
<u>y = -1</u>

Step-by-step explanation:

First, we must know the equations given:

2x + 3y = 1
3x + y = 5

Following Crammer's Rule, we have the matrix form:

$\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]$

Now we solve using the determinants:

$x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2$

$y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1$

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

2x + 3y = 1
2(2) + 3(-1)= 1
4 - 3 = 1
1 = 1

And, with the second equation:

3x + y = 5
3(2) + (-1) = 5
6 - 1 = 5
5 = 5

4 0

2 years ago

Plz help The population of a specific species of a nocturnal mammal is decreasing at a rate of 3.5%/year. The graph models the n

jeka94

Answer:

It is a hard question i can help tho.

Step-by-step explanation:

5 0

3 years ago