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Vilka [71]
3 years ago
11

A 30-m tall building casts a shadow. The distance from the top of the building to the tip of the shadow is 32 m. Find the length

of the shadow. If necessary, round your answer to the nearest tenth.
Mathematics
1 answer:
zepelin [54]3 years ago
7 0

Answer: 11.1 m

Step-by-step explanation:

This can be solved by the Pythagorean theorem where;

c² = a² + b²

c is the hypotenuse which is the distance from the top of the building to the tip of the shadow.

a is the height

c is the length

32² = 30² + b²

b² = 32² - 30²

b² = 1,024 - 900

b² = 124

b = √124

b = 11.1 m

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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

8 0
3 years ago
Shandra's parents put $1,500 in her bank account for college tuition. Given an interest
Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

8 0
3 years ago
Find the missing coefficient. (5d-7)(5d-6)=25d^2+__d+42
krok68 [10]

Answer:

-65

Step-by-step explanation:

(5d-7)(5d-6)

5d(5d)+5d(-6)-7(5d)-7(-6)

25d^2+-30d+-35d+42

25d^2+-65d+42

The missing coefficient of d is -65.

6 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH MY ANGLE PAIRS...If you help i will give your 5 stars, a thanks, and a brainliest :)
belka [17]

Answer:

x = 15

Step-by-step explanation:

9x - 5 = 7x + 25 ......(vertical angle and corresponding angle )

9x - 7x = 25 + 5

2x = 30

x = 30 / 2

x = 15

6 0
2 years ago
A bicycle training wheel has a radius of 3 inches. The bicycle wheel has a radius of 10 inches. Approximately how much smaller,
zhuklara [117]

Answer:

285.74\text{ inch}^2

Step-by-step explanation:

GIVEN: A bicycle training wheel has a radius of 3\text{ inch}. The bicycle wheel has a radius of 10\text{ inch}.

TO FIND: Approximately how much smaller, in square inches, is the area of the training wheel than the area of the regular wheel.

SOLUTION:

radius of bicycle wheel =10\text{ inch}

area of bicycle wheel =\pi r^2

putting value,

area of bicycle wheel =3.14\times10^2

                                    =314\text{ inch}^2

radius of bicycle training wheel =3\text{ inch}

area of bicycle training wheel =\pi r^2

putting value,

area of bicycle wheel =3.14\times3^2

                                    =28.26\text{ inch}^2

difference in area of bicycle wheel and bicycle training wheel

=\text{area of bicycle wheel}-\text{area of bicycle training wheel}

=314-28.26

=285.74\text{ inch}^2

hence the training wheel is 285.74\text{ inch}^2 smaller than regular wheel

4 0
3 years ago
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