Is it possible to have a triangle with angles of 62, 104 and 110
2 answers:
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Answer:
280.6
Step-by-step explanation:
<u>find the length of each side:</u>
you're only given h = 9 m
the formula for h is , where a is a side length of the hexagon
plug h in:
a ≈ 10.392 m
<u>now find the area of a triangle:</u>
1/2 (10.392)(9) = 46.764
multiply that by 6 and you get 280.584 ≈ 280.6
you can also use the regular hexagon area formula instead of finding the individual area and multiplying by 6:
Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
ANSWER
56.1 square inches.
EXPLANATION
The area of this triangle can be calculated using the formula,
where a=8in.
b=14.2 in.
and C=99° is the included angle.
We plug in these values to obtain,
We round to the nearest tenth to get,
56.1 sq. inch