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crimeas [40]
3 years ago
14

A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find

the velocity of the object when it is 30.0 m above the ground. b.Find the total distance the object travels during the fall.
Mathematics
1 answer:
Lyrx [107]3 years ago
4 0
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2 
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
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A blackjack player at a Las Vegas casino learned that the house will provide a free room if play is for four hours at an average
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Answer:

a) player’s expected payoff is $ 240

b) probability the player loses $1000 or more is 0.1788

c)  probability the player wins is 0.3557

d) probability of going broke is 0.0594

Step-by-step explanation:

Given:

Since there are 60 hands per hour and the player plays for four hours then the sample size is:

n = 60 * 4 = 240

The player’s strategy provides a probability of .49 of winning on any one hand so the probability of success is:

p = 0.49

a)

Solution:

Expected payoff is basically the expected mean

Since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand. So

Expected loss =  μ

                        = ∑ x P(x)

                        = 50 * P(win) - 50 * P(lose)

                        = 50 * P(win) + (-50) * (1 - P(win))

                         = 50 * 0.49 - 50 * (1 - 0.49)

                        = 24.5 - 50 ( 0.51 )

                        = 24.5 - 25.5

                        = -1

Since n=240 and expected loss is $1 per hand then the expected loss in four hours is:

240 * 1 = $ 240

b)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 110.5 - 117.6 / √117.6(1-0.49)

  = −7.1/√117.6(0.51)

  = −7.1/√59.976

  = −7.1/7.744417

  =−0.916789

Here the player loses 1000 or more when he loses at least 130 of 240 hands so the wins is 240-130 = 110

Using normal probability table:

P(X≤110) = P(X<110.5)

             = P(Z<-0.916)

             = 0.1788

c)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 120.5 - 117.6 / √117.6(1-0.49)

  = 2.9/√117.6(0.51)

  = 2.9/√59.976

  = 2.9/7.744417

  =0.374463

Here the player wins when he wins at least 120 of 240 hands

Using normal probability table:

P(X>120) = P(X>120.5)

              = P(Z>0.3744)  

             =  1 - P(Z<0.3744)

             = 1 - 0.6443

             = 0.3557

d)

Player goes broke when he loses $1500

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 105.5 - 117.6 / √117.6(1-0.49)

  = -12.1/√117.6(0.51)

  = -12.1/√59.976

  = -12.1/7.744417

  =−1.562416

Here the player loses 1500 or more when he loses at least 135 of 240 hands so the wins is 240-135 = 105

Using normal probability table:

P(X≤105) = P(X<105.5)

             = P(Z<-1.562)

             = 0.0594

7 0
3 years ago
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