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2 years ago

Look at the set of quadrilaterals in the coordinate plane shown below.

1 answer:

3 0

The true statement about the set of quadrilaterals in the coordinate plane is B. Because quadrilateral ABCD can be reflected across the x -axis, then rotated 90° counterclockwise about the origin, and then dilated about the origin by a scale factor of 2 to obtain quadrilateral A'B'C'D' , then quadrilateral ABCD is similar to quadrilateral A'B'C'D'.

<h3>What is a quadrilateral?</h3>

A quadrilateral is a polygon having four sides, four angles, and four vertices.

In this case, because quadrilateral ABCD can be reflected across the x -axis, then rotated 90° counterclockwise about the origin, and then dilated about the origin by a scale factor of 2 to obtain quadrilateral A'B'C'D' , then quadrilateral ABCD is similar to quadrilateral A'B'C'D'.

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Find the simplest form of the general term for the sequence 11, 15, 19, 23, .

sveta [45]

The general term of an arithmetic sequence is
an = a1 +(n-1)d
Your sequence has first term a1 = 11 and common difference d = 4. Then the general term is
an = 11 + (n-1)*4

The simplest form of the general term is
an = 7 + 4n

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3 years ago

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Amar rakes leaves for his neighbors to earn money he earned 64 dollars after 8 hours of work

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Amar makes 8 dollars an hour for raking leaves
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2 years ago

Which of the following would not represent the phrase “the difference of two numbers divided by 5”?

Sophie [7]

It would be (a) because it is now
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2 years ago

In August 2003, 56% of employed adults in the United States reported that basic mathematical skills were critical or very import

Serga [27]

Answer:

Yes

Step-by-step explanation:

First, suppose that nothing has changed, and possibility p is still 0.56. It's our null hypothesis. Now, we've got Bernoulli distribution, but 30 is big enough to consider Gaussian distribution instead.

It has mean μ= np = 30×0.56=16.8

standard deviation s = √npq

sqrt(30×0.56×(1-0.56)) = 2.71

So 21 is (21-16.8)/2.71 = 1.5494 standard deviations above the mean. So the level increased with a ˜ 0.005 level of significance, and there is sufficient evidence.

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3 years ago

Use the relative frequency approach to estimate the probability of the event. In a certain class of students, there are 8 boys f

In-s [12.5K]

Answer:

$P\text{Student will be from Kenilworth}\approx 0.256$

Step-by-step explanation:

We have been given that that there are 8 boys from Wilmette, 5 girls from Kenilworth, 9 girls from Wilmette, 5 boys from Glencoe, 5 boys from Kenilworth and 7 girls from Glenoce.

The total number of students from Kenilworth is 5 boys plus 5 girls that is 10 students.

The total number of students in the class would be $8+5+9+5+5+7=39$

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P\text{Student will be from Kenilworth}=\frac{\text{Number of student from Kenilworth}}{\text{Total number of students}}$

$P\text{Student will be from Kenilworth}=\frac{10}{39}$

$P\text{Student will be from Kenilworth}=0.2564102564$

$P\text{Student will be from Kenilworth}\approx 0.256$

Therefore, the probability, that the student will be from Kenilworth, is approximately 0.256.

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3 years ago