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ddd [48]
4 years ago
9

Triangle 1 is transformed onto triangle 2 in the graph below.

Mathematics
2 answers:
Serhud [2]4 years ago
7 0

Answer:

C

Step-by-step explanation:

1) you have to rotate triangle 1

2) then bring it down vertically and slide it over

Vesna [10]4 years ago
4 0

Answer:

Its D on edg i took the test

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Find the domain of the following function:
Lelechka [254]
What's the function?
7 0
4 years ago
Based on data from a​ college, scores on a certain test are normally distributed with a mean of 1530 and a standard deviation of
harkovskaia [24]

Answer:

0.73% of the scores are greater than 2317.

14.46% of the scores are less than 1190.

38.23% of the scores are between 1351 and 1673.

Step-by-step explanation:

Problems of normally distributed samples can be solved using 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.

In this problem, we have that:

\mu = 1530, \sigma = 322

Find the percentage of scores greater than 2317.

This is 1 subtracted by the pvalue of Z when X = 2317. So:

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

Z = \frac{2317 - 1530}{322}

Z = 2.44

Z = 2.44 has a pvalue of 0.9927.

So 1-0.9927 = 0.0073 = 0.73% of the scores are greater than 2317.

Find the percentage of scores less than 1190.

This is the pvalue of Z when X = 1190. So:

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

Z = \frac{1190 - 1530}{322}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446.

So 14.46% of the scores are less than 1190.

Find the percentage of scores between 1351 and 1673.

This is the pvalue of Z when X = 1673 subtracted by the pvalue of Z when X = 1351. So

X = 1673

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

Z = \frac{1673- 1530}{322}

Z = 0.44

Z = 0.44 has a pvalue of 0.67

X = 1351

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

Z = \frac{1351- 1530}{322}

Z = -0.56

Z = -0.56 has a pvalue of 0.2877

So 0.67-0.2877 = 0.3823 = 38.23% of the scores are between 1351 and 1673.

4 0
3 years ago
HELPPPPPPPPPPPPPPPPPP MEEEEEEEE
solmaris [256]

Answer:D

Step-by-step explanation:

The equation is y=mx+b. Here the slope is 1/5 and b is 5

4 0
2 years ago
Read 2 more answers
Please help! I have to use the Pythagorean theorem for this but idk how to do that with four numbers
lana [24]

Answer:

Step-by-step explanation:

Remark

Simple answer: you can't. I mean that you can't try to use 4 numbers, but you can solve the problem. You are going to have to redraw the diagram on another sheet of paper. Follow the directions below.

Directions for diagram extension.

Go to the right hand end of the 10 unit line.

Draw a line from the intersection point of the 10 unit line and 12 unit line

Draw this line so it is perpendicular to the 18 unit line. That will mean that the new line is parallel (and equal) to x

Mark the intersect point of the new line and the 18 unit line as B

Mark the intersect point of the 18 point line and the 12 unit line as C

Given and constructed

BC = 18 - 10 = 8

BC is one leg of the Pythagorean triangle.

The new x is the other leg of the Pythagorean triangle.

12 is the hypotenuse.

Formula

x^2 + 8^2 = 12^2

x refers to the new x which is equal to the given x

Solution

x^2 + 64 = 144              Subtract  64 from both sides

x^2 +64 - 64 = 144-64  Combine

x^2 = 80                        Break 80 down.

x^2 = 4 * 4 * 5               Take the square root of both sides        

x = 4*sqrt(5)

Comment

If you want the area it is 4*sqrt(5)(10 + 18)/2 = 56*sqrt(5)

5 0
4 years ago
Mark is a professional race car driver. He has competed in 15 races and won 40% of them. How many races has Mark won?
Romashka [77]

Answer:

6 is the answer

Step-by-step explanation:

8 0
4 years ago
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