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blondinia [14]
3 years ago
8

Of all the rectangles with a perimeter of 168 feet, find the dimension of the one with the largest area

Mathematics
1 answer:
dolphi86 [110]3 years ago
3 0
<span>The rectangle with the largest area with a given perimeter will be a square - so the sides will be equal. So we need to find length of side, L, such that 4*L=168. L = 168/4 L=42. So the dimensions of the rectangle that maximizes the area with a perimiter of 168 feet are: 42 feet by 24 feet.</span>
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Answer:

(a) <em>X</em> = number of dealerships the customer needs to call before she finds a used red Miata car.

(b) <em>X</em> = 1, 2, 3, 4,...

(c) The distribution of <em>X</em> is, X\sim Geometric\ (p=0.28).

(d) The expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.

(e) The probability that she must call at most 4 dealerships is 0.7313.

(f) The probability that she must call 3 or 4 dealerships is 0.2497.

Step-by-step explanation:

(a)

The random variable <em>X</em> can be defined as the number of dealerships the customer needs to call before she finds a used red Miata car.

(b)

The values that the random variable <em>X</em> can assume are:

<em>X</em> = 1, 2, 3, 4,...

(c)

The random variable <em>X</em> is defined as the number of trials before the first success. Each trial is independent of the others. And each trial has a similar probability of success.

This implies that the random variable <em>X</em> follows a Geometric distribution.

The probability of success, i.e. the probability that any independent dealership will have the car is <em>p</em> = 0.28.

Thus, the distribution of <em>X</em> is, X\sim Geometric\ (p=0.28).

(d)

The expected value of a Geometric distribution is:

E(X)=\frac{1}{p}

Compute the expected number of dealerships would we expect her to call until she finds one that has the car as follows:

E(X)=\frac{1}{p}

         =\frac{1}{0.28}\\

         =3.57

Thus, the expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.

(e)

Compute the probability that she must call at most 4 dealerships as follows:

P (X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

              =\sum\limits^{4}_{x=1}{(1-0.28)^{x-1}\times 0.28}\\=0.28+0.2016+0.1452+0.1045\\=0.7313

Thus, the probability that she must call at most 4 dealerships is 0.7313.

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Compute the probability that she must call 3 or 4 dealerships as follows:

P (X = 3 or X = 4) = P (X = 3) + P (X = 4)

                            =[(1-0.28)^{2-1}\times 0.28]+[(1-0.28)^{4-1}\times 0.28]\\=0.14515+0.10451\\=0.24966\\\approx0.2497

Thus, the probability that she must call 3 or 4 dealerships is 0.2497.

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