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n200080 [17]
3 years ago
9

The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Mathematics
1 answer:
Nuetrik [128]3 years ago
8 0

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

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The median of Rae's data is 2.22 minutes, and the median of Doris' data is 2.14 minutes. The interquartile range for Rae is , an
JulsSmile [24]

Complete question :

Rae and Doris are training to swim a 200-meter freestyle race. The table lists their practice times during training camp.

Rae’s Times(minutes)-2.12, 2.01, 2.46, 2, 2.22, 2.31, 2.23

doris times(minutes)-2.32, 2.19, 2.26, 2.03, 2.11, 2.14, 2.07

The median of Raes data is(drop down box answers, 2,2.03,2.14,2.22)minutes,and the median of doris data is(drop down box answers, 2,2.03,2.14,2.22)minutes. The interquartile range for rae is(drop down box answers, 0.01,0.13,0.26,0.30), and the interquartile range for doris is(drop down box answers, 0.02,0.06,0.18,0.19). The two data sets overlap(drop down box answers,very little, a lot

Answer:

Rae's median = 2.22

Rae's IQR = 0.30

DORIS median = 2.14

Doris IQR = 0.19

Step-by-step explanation:

Given :

Rae's time :

Ordered data : 2, 2.01, 2.12, 2.22, 2.23, 2.31, 2.46

Median = 1/2(n+1)th term ; n = 7

Median = 1/2(8) = 4th term

Median = 2.22

Q1 = 1/4(8)th term

Q1 = 2nd term = 2.01

Q3 = 3/4(n + 1)th term

Q3 = 3/4(8)th term

Q3 = 6th term = 2.31

Interquartile range = Q3 - Q1 = 2.31 - 2.01 = 0.30

DORIS Data:

Ordered data: 2.03, 2.07, 2.11, 2.14, 2.19, 2.26, 2.32

Median = 1/2(n+1)th term ; n = 7

Median = 1/2(8) = 4th term

Median = 2.14

Q1 = 1/4(8)th term

Q1 = 2nd term = 2.07

Q3 = 3/4(n + 1)th term

Q3 = 3/4(8)th term

Q3 = 6th term = 2.26

Interquartile range = Q3 - Q1 = 2.26 - 2.07 = 0.19

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