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White raven [17]
2 years ago
10

You're flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 201 km south and 194 km east. Mt. Rainier is locat

ed approximately 56 km east and 40 km south of JBLM. If you are flying at a constant speed of 800 km/hr, how long after you depart JBLM will you be the closest to Mt. Rainier?
Mathematics
1 answer:
elixir [45]2 years ago
6 0

Answer:

The answer is "0.0846276476".

Step-by-step explanation:

Let all the origin(0,0) of JBLM be  

Let the y-axis be north along with the vector j unit.  

But along the + ve x-axis is east, and all along with the vector unit i.  

And at (194,-201) that undisclosed position is  

Mt. Ris (56,-40) at  

Let the moment it gets close to the Mt. rainier bet.  

Oh, then,  

If we know, the parallel to the direction from the point was its nearest one to a path to a point,

Calculating slope:

\to m=\frac{y_2-y_1}{x_2-x_1}\\

points:  (0,0) and (194,-201)

\to m=\frac{-201 -0 }{194-0}\\\\\to m=\frac{-201}{194}\\\\\to m=-1.03

equation of the line:

\to y= mx+c\\\\\to y= -1.03\ x+c

when the slope is perpendicular= -\frac{1}{m}

                                                     = - \frac{1}{ -1.03}\\\\= \frac{1}{ 1.03}\\\\= 0.97

perpendicular equation:

\to y-(-40)=0.97  \times (x-56) \\\\\to y+40=0.97  \times (x-56) \\\\

Going to solve both of the equations to have the intersection point,  

We get to the intersect level in order to be at

(47.16,-48.5748)

so the distance from origin:

= \sqrt{(47.16^2+48.5748^2)}\\\\= \sqrt{2224.0656 + 2359.5112}\\\\=\sqrt{4583.5768}\\\\=67.7021181

\to time =\frac{distance }{speed}

= \frac{67.7021181}{800}\\\\= 0.0846276476

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kotegsom [21]

Answer:

9 19/21

Step-by-step explanation:

2 and 4/7 + 7 3/9 or 7 1/3. If you find common denominator its 21. So 2 and 12/21 and 7 7/21. If you add that would equal 9 19/21.

Hope this helps :)

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2 years ago
Find the area of a trapezoid with bases that measure 9.2 CM and 7.7 cm and height 6.8 cm
maria [59]
\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\
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a=9.2\\
b=7.7\\
h=6.8
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3 years ago
Read 2 more answers
Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F(x) =
Troyanec [42]

Answer:

a) P (x <= 3 ) = 0.36

b) P ( 2.5 <= x <= 3  ) = 0.11

c) P (x > 3.5 ) = 1 - 0.49 = 0.51

d) x = 3.5355

e) f(x) = x / 12.5

f) E(X) = 3.3333

g) Var (X) = 13.8891  , s.d (X) = 3.7268

h) E[h(X)] = 2500

Step-by-step explanation:

Given:

The cdf is as follows:

                           F(x) = 0                  x < 0

                           F(x) = (x^2 / 25)     0 < x < 5

                           F(x) = 1                   x > 5

Find:

(a) Calculate P(X ≤ 3).

(b) Calculate P(2.5 ≤ X ≤ 3).

(c) Calculate P(X > 3.5).

(d) What is the median checkout duration ? [solve 0.5 = F()].

(e) Obtain the density function f(x). f(x) = F '(x) =

(f) Calculate E(X).

(g) Calculate V(X) and σx. V(X) = σx =

(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].

Solution:

a) Evaluate the cdf given with the limits 0 < x < 3.

So, P (x <= 3 ) = (x^2 / 25) | 0 to 3

     P (x <= 3 ) = (3^2 / 25)  - 0

     P (x <= 3 ) = 0.36

b) Evaluate the cdf given with the limits 2.5 < x < 3.

So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3

     P ( 2.5 <= x <= 3  ) = (3^2 / 25)  - (2.5^2 / 25)

     P ( 2.5 <= x <= 3  ) = 0.36 - 0.25 = 0.11

c) Evaluate the cdf given with the limits x > 3.5

So, P (x > 3.5 ) = 1 - P (x <= 3.5 )

     P (x > 3.5 ) = 1 - (3.5^2 / 25)  - 0

     P (x > 3.5 ) = 1 - 0.49 = 0.51

d) The median checkout for the duration that is 50% of the probability:

So, P( x < a ) = 0.5

      (x^2 / 25) = 0.5

       x^2 = 12.5

      x = 3.5355

e) The probability density function can be evaluated by taking the derivative of the cdf as follows:

       pdf f(x) = d(F(x)) / dx = x / 12.5

f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:

         E(X) = integral ( x . f(x)).dx          limits: - ∞ to +∞

         E(X) = integral ( x^2 / 12.5)    

         E(X) = x^3 / 37.5                    limits: 0 to 5

         E(X) = 5^3 / 37.5 = 3.3333

g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:

         Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2          limits: - ∞ to +∞

         Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2    

         Var(X) = x^4 / 50 | - (3.3333)^2                         limits: 0 to 5

         Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891

         s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268

h) Find the expected charge E[h(X)] , where h(X) is given by:

          h(x) = (f(x))^2 = x^2 / 156.25

  The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:

         E(h(X))) = integral ( x . h(x) ).dx          limits: - ∞ to +∞

         E(h(X))) = integral ( x^3 / 156.25)    

         E(h(X))) = x^4 / 156.25                       limits: 0 to 25

         E(h(X))) = 25^4 / 156.25 = 2500

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