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4 years ago

Enter the data to create a histogram. Then check all that apply.

Mathematics

2 answers:

fomenos4 years ago

8 0

Answer:

Enter the data to create a histogram. Then check all that apply.

A 1-column table with 10 rows. Column 1 is labeled runs with entries 1, 7, 1, 6, 1, 2, 8, 6, 8, 0.

The histogram has 1 cluster.

The histogram has 2 clusters.

The histogram has 1 peak.

The histogram has 4 peaks.

The histogram has 0 frequency from 4-5.Step-by-step explanation:

Ksju [112]4 years ago

4 0

Answer:

B,C,E

Step-by-step explanation:

The histogram has 2 cluster

The histogram 1 peak

The histogram has 0 frequency from 4-5

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Gemiola [76]

It is 50. the angles on the side of ABC are the same as on ACD, where they are labelled. When you add the two angles on the other side together, the answer is 50 degrees.

5 0

3 years ago

Square RSTU is inscribed in circle P. Given the coordinates for the vertices of the square, find the equation of circle P.

svlad2 [7]

Answer:

$(x-2)^2+y^2=20$

Step-by-step explanation:

We want to find the equation of a circle P, and we know that the square RSTU is inscribed in the circle, this means that if we replace the points:

R(0,4) S(6,2) T(4,-4) and U(-2,-2) the equation must be verified.

Then, we have to replace the points in each option to see which one is the correct answer.

Option A: $(x-2)^2+y^2=68$

R(0,4):

$(x-2)^2+y^2=68\\(0-2)^2+4^2=68\\(-2)^2+16=68\\4+16=68\\20\neq 68$

We can see that R doesn't belong to the equation of the circle.

We could replace the rest of the points in the equation, but since we already know that R does not belong to the circle, we already know that it is not the right option.

Option B: $(x+2)^2+y^2=20$

R(0,4):

$(x+2)^2+y^2=20\\(0+2)^2+4^2=20\\2^2+4^2=20\\4+16=20\\20=20$

The point R(0,4) verifies the equation.

S(6,2):

$(x+2)^2+y^2=20\\(6+2)^2+2^2=20\\8^2+4^2=20\\64+16=20\\80\neq 20$

The point S doesn't belong to the equation of the circle, then we already know that this isn't the correct option.

Option C: $(x-2)^2+y^2=20$

R(0,4):

$(x-2)^2+y^2=20\\(0-2)^2+4^2=20\\(-2)^2+16=20\\4+16=20\\20=20$

The point R verifies the equation.

S(6,2):

$(x-2)^2+y^2=20\\(6-2)^2+2^2=20\\4^2+4=20\\16+4=20\\20=20$

The point S verifies the equation.

T(4,-4):

$(x-2)^2+y^2=20\\(4-2)^2+(-4)^2=20\\2^2+16=20\\4+16=20\\20=20$

The point T verifies the equation.

U(-2,-2):

$(x-2)^2+y^2=20\\(-2-2)^2+(-2)^2=20\\(-4)^2+4=20\\16+4=20\\20=20$

The point U verifies the equation.

<em>We can see that the four points RSTU verifies the equation. This means that the square RSTU is inscribed in circle P whose equation is: </em> $(x-2)^2+y^2=20$

<em>We can see that the graph of the equation verifies the answer.</em>