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polet [3.4K]
3 years ago
13

The greenhouse club is purchasing seed for the lawn in the school courtyard. The club needs to determine how much to buy. Unfort

unately, the club meets after school, and students are unable to find a custodian to unlock the door. Anthony suggests they just use his school map to calculate the area that will need to be covered in seed. He measures the rectangular area on the map and finds the length to be 10 inches and the width to be 6 inches. The map notes the scale of 1 inch representing 7 feet in the actual courtyard. What is the actual area in square feet?
Mathematics
1 answer:
kiruha [24]3 years ago
7 0

Answer:

The answer is 2940ft^2

Step-by-step explanation

First convert length and width to feet using the value given (7ft=1inch).

10*7= 70

6*7= 42

Multiply these values for the actual area.

42*70= 2940

The actual area of the lawn is 2940ft^2

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Leokris [45]

Given:

The data values are

11, 12, 10, 7, 9, 18

To find:

The median, lowest value, greatest value, lower quartile, upper quartile, interquartile range.

Solution:

We have,

11, 12, 10, 7, 9, 18

Arrange the data values in ascending order.

7, 9, 10, 11, 12, 18

Divide the data in two equal parts.

(7, 9, 10), (11, 12, 18)

Divide each parenthesis in 2 equal parts.

(7), 9, (10), (11), 12, (18)

Now,

Median = \dfrac{10+11}{2}

            = \dfrac{21}{2}

            = 10.5

Lowest value = 7

Greatest value = 18

Lower quartile = 9

Upper quartile = 12

Interquartile range (IQR) = Upper quartile - Lower quartile

                                        = 12 - 9

                                        = 3

Therefore, median is 10.5, lowest value is 7, greatest value is 18, lower quartile 9, upper quartile 12 and interquartile range is 3.

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9514 1404 393

Answer:

  1. Angle 1 = 139°
  2. Angle 2 = 41°
  3. x = 29; exterior angle = 131°

Step-by-step explanation:

These problems let you make use of the fact that the sum of the remote interior angles is equal to the exterior angle.

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1. 53° +86° = ∠1

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2. ∠2 +92° = 133°

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The exterior angle is ...

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