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SOVA2 [1]
3 years ago
7

The head of the math department compared the scores of students in two classes on the Chapter 2 test. He found that the mean sco

re of both classes was the same. He also found that there was greater variability in the scores of Ms. Biren’s students than in the scores of Mr. Hayes’s students. Which quality of the data sets would lead the department head to that conclusion? The median score in Ms. Biren’s class is greater than the median score in Mr. Hayes’s class.The median score in Ms. Biren’s class is less than the median score in Mr. Hayes’s class.The range of scores in Ms. Biren’s class is greater than the range of scores in Mr. Hayes’s class.The range of scores in Ms. Biren’s class is less than the range of scores in Mr. Hayes’s class.
Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
5 0
Given:
Mean of Ms. Biren's students score are the same with the mean of Mr. Haye's students scores.

Ms. Biren's students scores have greater variability than Mr. Haye's students scores.

The quality of the data set that would lead the department head to that conclusion is:

<span>The range of scores in Ms. Biren’s class is greater than the range of scores in Mr. Hayes’s class.</span>
arlik [135]3 years ago
3 0

Answer:

he range of scores in Ms. Biren’s class is greater than the range of scores in Mr. Hayes’s class.

Step-by-step explanation:

The range is one measure of variability.  The larger the range is, the greater the distance is between the lowest and the highest values; this means the data with a greater range is more variable, and the data with a lower range is less variable.

Since Ms. Biren's class has a higher range, this means her class has greater variability.


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Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of k, where k
tresset_1 [31]

Answer:

a) Similar

b) Perimeter of rectangle 2 is k times the Perimeter of rectangle 1 (Proved Below)

c) Area of rectangle 2 is k^2 times the Area of rectangle 1 (Proved Below)

Step-by-step explanation:

Given:

Rectangle 1 has length = x

Rectangle 1 has width = y

Rectangle 2 has length = kx

Rectangle 2 has width = ky

(a) Are Rectangle 1 and Rectangle 2 similar? Why or why not?

Rectangle 1 and Rectangle 2 are similar because the angles of both rectangles are 90° and the sides of Rectangle 2 is k times the sides of Rectangle 1. So sides of both rectangles is equal to the ratio k.

(b) Write a paragraph proof to show that the perimeter of Rectangle 2 is k times the perimeter of Rectangle 1.

Perimeter of Rectangle = 2*(Length + Width)

Perimeter of Rectangle 1 = 2*(x+y) = 2x+2y

Perimeter of Rectangle 2 = 2*(kx+ky) = 2kx + 2ky

                                          = k(2x+2y)

                                          = k(Perimeter of Rectangle 1)

Hence proved that Perimeter of rectangle 2 is k times the perimeter of rectangle 1.

(c) Write a paragraph proof to show that the area of Rectangle 2 is k^2 times the area of Rectangle 1.

Area of Rectangle = Length * width

Area of Rectangle 1 = x * y

Area of Rectangle 2 = kx*ky

                                  = k^2 (xy)

                                  = k^2 (Area of rectangle 1)

Hence proved that area of rectangle 2 is k^2 times the area of rectangle 1.

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3 years ago
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Answer:

c 2

Step-by-step explanation:

h/3 + h/6 =1

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