Median of the data distribution given in the histogram using the cumulative frequency is 8.
The class width = 1
The values of the data are,
4, 1, 1, 1, 1, 3, 4, 4, 6.
Cumulative frequencies = 4, 5, 6, 7, 8, 11, 15, 19, 25
n = 25 and n/2 = 12.5
Median lies on the 12.5th observation.
12.5th observation lies in the class where cumulative frequency is 15.
Median class is 7.5 - 8.5.
l, lower limit of median class = 7.5
n = 25
f = frequency of median class = 4
cf, cumulative frequency of class preceding the median class = 11
h, class size = 1
Median = l + [(n/2 - cf) / f] h
= 7.5 + [(25/2 - 11) / 4] 1
= 7.875 ≈ 8
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Answer:
The first one
Step-by-step explanation:
Let ℓ equal length, w equal width, and P equal perimeter.
Use the formula 2ℓ + 2w = P.
ℓ = 2w - 5
2ℓ + 2w = 62
Substitute what you do know for what you don't, and solve.
2(2w-5) + 2w = 62
4w - 10 + 2w = 62
6w - 10 = 62
6w = 72
w = (72/6)
w = 12
The answer is 12 feet.
The width of the deck is 12.0 feet.
To solve this problem, we can set up an equation using information given. Let's say the width of the deck is x feet. According to the problem, the length of the deck is 5 less than twice the width, so the length would be 2x - 5.
The perimeter of a rectangle is given by the formula P = 2(l + w), where P is the perimeter, l is the length, and w is the width. Therefore, we can write the equation as:
62 = 2(x + (2x - 5))
Simplify and solve for x:
62 = 6x - 10
72 = 6x
x = 12
Rounded to the nearest tenth, the width of the deck is 12.0 feet.
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than one foot