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____ [38]
2 years ago
15

PLZZZZZZZZZ help its on the screenshot

Mathematics
2 answers:
Komok [63]2 years ago
6 0

Answer:

the answer is 40%

Step-by-step explanation:

liraira [26]2 years ago
6 0
Yea I also got 40% but my teacher says it’s wrong :(
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Polygon ABCD is plotted on a coordinate plane and then rotated 90° clockwise about point C to form polygon A′B′C′D′. Match each
horrorfan [7]

The vertices of ABCD after 90 degrees clockwise rotation about point C are: A' (0,6), B' (3,7), C' (4,6) and D' (4,3)

<h3>How to match the vertices of the polygon?</h3>

The image of the polygon ABCD is not given; however, the question can still be answered because the coordinates are known.

The vertices of polygon ABCD are given as:

A = (4, 6)

B = (5, 3)

C = (4, 2)

D = (1, 2)

The rule of rotation about point C is:

(x,y) = (a + b - y, x + b - a)

Where:

(a, b) = (4, 2) --- the point of rotation.

So, we have:

(x,y) = (4 + 2 - y, x + 4 - 2)

(x,y) = (6 - y, x + 2)

The above means that:

A' = (6 - 6, 4 + 2) = (0,6)

B' = (6 - 3, 5 + 2) = (3,7)

C' = (6 - 2, 4 + 2) = (4,6)

D' = (6 - 2, 1 + 2) = (4,3)

Hence, the image of the rotation and their vertices (i.e. coordinates) are:

A' = (0,6)

B' =  (3,7)

C' = (4,6)

D' =  (4,3)

Read more about rotation at:

brainly.com/question/4289712

#SPJ1

8 0
1 year ago
If 75 is 25% of a value, what is that value?
sertanlavr [38]
75÷25%
25%=0.25
75÷0.25
=300
Your answer is 300
7 0
3 years ago
The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p
VLD [36.1K]

Answer:

3.13

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

\chi^2_{1-\frac{\alpha}{2}}= 51.770

Chi square critical value for upper tail =

\chi^2_{\frac{\alpha}{2}}= 81.085

80% confidence interval:

\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}

Putting values, we get,

=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0
3 years ago
Please help with this problem!!
hodyreva [135]

The answer is A because the two lines are equal

8 0
3 years ago
9.
Elena L [17]

<u>Given</u>:

The distance (in kilometers) Dora hiked is modeled as a function of time.

We need to determine the average rate of change in distance hiked, measured in kilometers per hour, between 8:30 am to 1:30 pm.

<u>Average rate of change:</u>

Let us write the time and the distance hiked in coordinates for the time 8:30 am and 1:30 pm.

Thus, the coordinates are (8.30, 4) and (1,12)

The average rate of change can be determined using the formula,

m=\frac{y_2-y_1}{x_2-x_1}

Substituting the points (8.30, 4) and (1.30,12), we get;

m=\frac{12-4}{1.30-8.30}

m=\frac{8}{-7}

m=-1.14

Rounding off to the nearest whole number, we get;

m=-1

Therefore, the average rate of change is -1.

7 0
3 years ago
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