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NARA [144]
2 years ago
13

: West Gorham High School is to be located at the population center of gravity of three communities: Westbrook, population 16,00

0; Scarborough, population 22,000; and Gorham, population 36,500. Westbrook is located at 43.6769˚N, 70.3717˚W; Scarborough is located at 43.5781˚N, 70.3222˚W; and Gorham is located at 43.6795˚N, 70.4447˚W.
 a. Where should West Gorham High School be located?
 b. If only two pieces of adequate land are available for sale: Baker’s Field at 43.6784˚N, 70.3827˚W; or L onesome Acres at 43.5119˚N, 70.3856˚W, using rectilinear distances, which is closer to the site located in part (a)?
Mathematics
1 answer:
xenn [34]2 years ago
5 0

Answer:

  (a) 43.6490°N, 70.3928°W

  (b) Baker's Field

Step-by-step explanation:

The locations are all within a 7-mile square, so we will make the assumption of a flat Earth with parallel longitude lines. Then the "central" location will be the weighted average of the latitude and longitude coordinates.

<h3>(a)</h3>

Total population is 16000 +22000 +36500 = 74,500. For purposes of the arithmetic, we will subtract 43 from latitude and 70 from longitude, then multiply each fraction by 10. Similarly, we will divide each population by 1000.

The weighted average location is found from ...

  (16/74.5)(6.769, 3.717) +(22/74.5)(5.781, 3.222) +(36.5/74.5)(6.795, 4.447)

  = (1.4537, 0.7983) +(1.7071, 0.9515) +(3.3291, 2.1787) = (6.490, 3.928)

The center of gravity is about 43.6490°N, 70.3928°W.

__

<h3>(b)</h3>

The coordinate differences from the center of gravity are (in (°lat, °long)) ...

  Baker's Field: (0.0294, -0.0101)

  Lonesome Acres: (-0.1371, -.0072)

Clearly, Baker's field is closer to the site of part (a).

_____

<em>Additional comment</em>

Distance calculations could be refined further by considering that a degree of longitude at this latitude represents about 0.72 times the distance of a degree of latitude. In this problem, the location of Lonesome Acres has a latitude difference that is several times any of the other coordinate differences, so it is clearly much farther away from the desired location.

The calculations done here are a sort of first-level approximation. The least-significant digits in the coordinates represent a distance of less than 50 feet. If you really want accuracy to that level, then the (approximately) spherical shape of the Earth needs to be accounted for. As a second-level approximation, some reference point needs to be identified, and coordinates in miles (or feet) need to be expressed in relation to that reference. The best accuracy would be obtained using spherical trigonometry to identify the distances.

We expect any school that is constructed to be somewhat larger than 50 feet in any direction, so accuracy to the 4th decimal place in these locations is probably not required--except to satisfy some answer-checker.

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Serggg [28]

Answer:

k = 1

P(x > 3y) = \frac{2}{3}

Step-by-step explanation:

Given

f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }  & { \text 0, { elsewhere. } } \end{array} \right.

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1

Where

{ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }

Substitute values for the interval of x and y respectively

So, we have:

\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1

Isolate k

k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1

Integrate y, leave x:

k \int\limits^2_{0} y {dx} \, [0,x/2]= 1

Substitute 0 and x/2 for y

k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1

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Integrate x

k * \frac{x^2}{2*2} [0,2]= 1

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Substitute 0 and 2 for x

k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1

k *[ \frac{4}{4} - \frac{0}{4} ]= 1

k *[ 1-0 ]= 1

k *[ 1]= 1

k = 1

Solving (b): P(x > 3y)

We have:

f(x,y) = k

Where k = 1

f(x,y) = 1

To find P(x > 3y), we use:

\int\limits^a_b \int\limits^a_b {f(x,y)}

So, we have:

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0  dxdy

Integrate x leave y

P(x > 3y) = \int\limits^2_0  x [0,y/3]dy

Substitute 0 and y/3 for x

P(x > 3y) = \int\limits^2_0  [y/3 - 0]dy

P(x > 3y) = \int\limits^2_0  y/3\ dy

Integrate

P(x > 3y) = \frac{y^2}{2*3} [0,2]

P(x > 3y) = \frac{y^2}{6} [0,2]\\

Substitute 0 and 2 for y

P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}

P(x > 3y) = \frac{4}{6} -\frac{0}{6}

P(x > 3y) = \frac{4}{6}

P(x > 3y) = \frac{2}{3}

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2 years ago
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Maksim231197 [3]

Answer:

7)  C) PQR ≈ TSR

8) B)  3

9) D) 102 ft

10) D) 21

11) C) 45

Step-by-step explanation:

7) C) PQR ≈ TSR

8) 2 * 6 = 12

so 4x = 12

      x = 3

9)

24/20 = x/85

x = 24 * 85 / 20

x = 202 ft

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QT/35 = 24/40

QT/35 = 3/5

QT = 35 * 3 / 5

QT = 21

11)

ΔKLM ≈ΔXYZ

9/4 = 2.25

So

XY = 9 ---> given

YZ = 7 * 2.25 = 15.75

XZ = 9 * 2.25  = 20.25

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7 0
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Nataly_w [17]

Answer: D. Predictive validity

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3 years ago
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balu736 [363]

Answer:

∠4 and ∠3

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Step-by-step explanation:

A pair of angles is said to be supplementary when the 2 angles add up to give 180°.

Two right angles add up to give 180°. Also, if we have two angles on a straight line, their sum = 180°, according to the linear pair property.

Thus, since m< 3 is given as 90°, m<4 = 90°.

m<4 + m<3 = 180°.

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