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bogdanovich [222]
3 years ago
5

Estimate the perimeter of the figure to the nearest whole number.

Mathematics
1 answer:
Paha777 [63]3 years ago
4 0

Answer:

The perimeter (to the nearest integer) is 9.

Step-by-step explanation:

The upper half of this figure is a triangle with height 3 and base 6.  If we divide this vertically we get two congruent triangles of height 3 and base 3.  Using the Pythagorean Theorem we find the length of the diagonal of one of these small triangles:  (diagonal)^2 = 3^2 + 3^2, or (diagonal)^2 = 2*3^2.

Therefore the diagonal length is (diagonal) = 3√2, and thus the total length of the uppermost two sides of this figure is 6√2.

The lower half of the figure has the shape of a trapezoid.  Its base is 4.  Both to the left and to the right of the vertical centerline of this trapezoid is a triangle of base 1 and height 3; we need to find the length of the diagonal of one such triangle.  Using the Pythagorean Theorem, we get

(diagonal)^2 = 1^2 + 3^2, or 1 + 9, or 10.  Thus, the length of each diagonal is √10, and so two diagonals comes to 2√10.

Then the perimeter consists of the sum 2√10 + 4 + 6√2.

which, when done on a calculator, comes to 9.48.  We must round this off to the nearest whole number, obtaining the final result 9.

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P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda

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Number of these intervals in which exactly one car arrives is given by

0.2599\times 18=4.6798

We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

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P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

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Step-by-step explanation:

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Time Period = 5 \frac12\ hr

Now we can say that;

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