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m_a_m_a [10]
3 years ago
11

A flat rectangular piece of aluminum has a perimeter of 70 inches. The length is 11 inches longer than the width. Find the width

. A. 34 inches B. 35 inches C. 23 inches D. 12 inches
Mathematics
2 answers:
qwelly [4]3 years ago
8 0

Answer:

The answer is the option D

12\ inches

Step-by-step explanation:

we know that

The perimeter of a rectangle is equal to

P=2L+2W

where

L is the length side of the rectangle

W is the width side of the rectangle

In this problem we have

P=70\ in

so

70=2L+2W ------> equation A

L=W+11 ------> equation B

Substitute equation B in equation A and solve for W

70=2[W+11]+2W

70=2W+22+2W

4W=70-22

W=48/4

W=12\ in

rodikova [14]3 years ago
5 0

perimeter = 2L +2w

L = w+11

70 = 2(w+11) +2w

70 = 2w+22+2w

70= 4w + 22

48 = 4w

w=48/4 = 12

width = 12

length = 12+11 = 23

2x12 = 24

2x23 = 46

46+24 = 40

 length = 23 inches, width = 12 inches

Answer is D

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Marina CMI [18]

Answer:

The median of A is the same as the median of B.

The interquartile range of B is greater than the interquartile range of A.

Step-by-step explanation:

Given that:

A = number of runs allowed in first 9 games

A = {1, 4, 2, 2, 3, 1, 1, 2, 1}

Rearranging A : 1, 1, 1, 1, 2, 2, 2, 3, 4

Median A = 1/2(n + 1) th term

Median A = 1/2(10) = 5th term = 2

Q1 of A = 1/4(10) = 2.5th term = (1 + 1)/ 2 = 1

Q3 of A = 3/4(10) = 7.5th term = (2+3)/2 = 2.5

Interquartile range = Q3 - Q1 = 2.5 - 1 = 1.5

Number of runs allowed in 10th game = 9

B = {1, 4, 2, 2, 3, 1, 1, 2, 1, 9}

Rearranging B = 1, 1, 1, 1, 2, 2, 2, 3, 4, 9

Median A = 1/2(n + 1) th term

Median A = 1/2(11) = 5.5th term = (2+2)/2 = 2

Q1 of A = 1/4(11) = 2.75th tetm = (1 + 1)/ 2 = 1

Q3 of A = 3/4(11) = 8.25th term = (3+4)/2 = 3.5

Interquartile range = Q3 - Q1 = 3.5 - 1 = 2.5

Median A = 2 ; median B = 2

IQR B = 2.5 ; IQR A = 1.5 ; IQR B > IQR A

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Step-by-step explanation:

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