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Aleks04 [339]
3 years ago
7

EF is located at E (x,y) and F (7,-10) and was rotated around the origin. EF became E’F’ with coordinates E’ (4,5) and For (10,7

).What type of rotation was performed?
Mathematics
1 answer:
monitta3 years ago
5 0

Answer:

  90° CCW

Step-by-step explanation:

If we assume you intend F'(10, 7), point F has been transformed by the rule ...

  (x, y) ⇒ (-y, x)

The transformation rule (x, y) ⇒ (-y, x) represents a 90° CCW rotation.

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Step-by-step explanation:

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3 years ago
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Step-by-step explanation:

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3 0
3 years ago
Suppose that the national average for the math portion of the College Board's SAT is 515. The College Board periodically rescale
nasty-shy [4]

Answer:

a) 16% of students have an SAT math score greater than 615.

b) 2.5% of students have an SAT math score greater than 715.

c) 34% of students have an SAT math score between 415 and 515.

d) Z = 1.05

e) Z = -1.10

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the empirical rule.

Normal probability distribution

Problems of normally distributed samples are 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.

Empirical rule

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

\mu = 515, \sigma = 100

(a) What percentage of students have an SAT math score greater than 615?

615 is one standard deviation above the mean.

68% of the measures are within 1 standard deviation of the mean. The other 32% are more than 1 standard deviation from the mean. The normal probability distribution is symmetric. So of those 32%, 16% are more than 1 standard deviation above the mean and 16% more then 1 standard deviation below the mean.

So, 16% of students have an SAT math score greater than 615.

(b) What percentage of students have an SAT math score greater than 715?

715 is two standard deviations above the mean.

95% of the measures are within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. The normal probability distribution is symmetric. So of those 5%, 2.5% are more than 2 standard deviations above the mean and 2.5% more then 2 standard deviations below the mean.

So, 2.5% of students have an SAT math score greater than 715.

(c) What percentage of students have an SAT math score between 415 and 515?

415 is one standard deviation below the mean.

515 is the mean

68% of the measures are within 1 standard deviation of the mean. The normal probability distribution is symmetric, which means that of these 68%, 34% are within 1 standard deviation below the mean and the mean, and 34% are within the mean and 1 standard deviation above the mean.

So, 34% of students have an SAT math score between 415 and 515.

(d) What is the z-score for student with an SAT math score of 620?

We have that:

\mu = 515, \sigma = 100

This is Z when X = 620. So

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

Z = \frac{620 - 515}{100}

Z = 1.05

(e) What is the z-score for a student with an SAT math score of 405?

We have that:

\mu = 515, \sigma = 100

This is Z when X = 405. So

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

Z = \frac{405 - 515}{100}

Z = -1.10

3 0
4 years ago
What is the length of segment AE?
nexus9112 [7]

Answer:

AE=\frac{50}{3}\ units

Step-by-step explanation:

step 1

In the right triangle ABC

Applying the Pythagoras Theorem fin the hypotenuse AC

AC^{2} =AB^{2}+BC^{2}

substitute

AC^{2} =6^{2}+8^{2}

AC^{2} =100

AC =10\ units

step 2

we know that

If two figures are similar, then the ratio of its corresponding sides is equal

so

\frac{AB}{CD}=\frac{AC}{CE}

substitute and solve for CE

\frac{6}{4}=\frac{10}{CE}\\ \\CE=4*10/6\\ \\CE=\frac{20}{3}\ units

step 3

Find the length of segment AE

AE=AC+CE

substitute the values

AE=10\ units+\frac{20}{3}\ units=\frac{50}{3}\ units

6 0
4 years ago
HOW MANY SIDES ARE ON THIS TRIANGLE NEED HELP ASAP AND I NEED SOMEONE WHI ACTUALLY KNOWS NOT SOME SECOND GRADER
sladkih [1.3K]
There would be four sides, three around and the bottom base <span />
8 0
4 years ago
Read 2 more answers
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