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Ulleksa [173]
3 years ago
14

A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th

at face. For example, a six is three times as probable as a two.
What is the probability of getting either a 5 or a 2 in one throw?


Hint: Let m be a distribution function for this experiment. If a six is three times as probable as a two, then in terms of the distribution function, m(6) = 3. m(2)
Mathematics
1 answer:
agasfer [191]3 years ago
8 0

Answer:

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

P(6) = 6×P(1)

P(T) = 21×P(1)

Where;

P(x) is the probability of getting number x on the dice.

P(T) is the total probability of obtaining any number

N(x) is the number of possible number x in terms of the distribution function.

P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

P(x) is directly proportional to N(x)

So, equation 1 becomes;

P(x) = N(x)/N(T) = P(x)/P(T) ....2

The probability of getting either a 5 or a 2 in one throw

P(2U5) = (P(2) + P(5))/P(T)

Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

P(2U5) = 7P(1)/21P(1)

P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

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Let
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y----------------> </span><span>feet of yarn of James

we know that
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x+y=4 2/8-------> x+y=17/4----> equation 1
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substitute 2 in 1
[13/5]+y=17/4------> y=(17/4)-(13/5)----> y=[5*17-4*13]/20
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the answer is
</span><span>james have 1 13/20 ft of yarn</span>
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Answer:

4.75 seconds

Step-by-step explanation:

Initial height of the ball from the ground = 10 feet

Upward initial velocity given = 55 feet per second

Value of g ( acceleration due to gravity) = 32.2 feet per s²

Motion of the ball:

The ball first goes vertically upwards, gravity decelerates the body and it momentarily comes to rest at a point in its upward trajectory, which is the point of maximum height. From this point, to the ground, the ball behaves as a freely dropped body.

Till the ball reaches its maximum height:

u = + 55

v = 0 ( final velocity is zero)

a = - 32.2 (since it is deceleration)

we know that <em>v = u + at </em>

⇒    0 = 55 - 32.2t

⇒  t = 1.7 s

Also , we have <em>s = ut + (1/2)at²</em>

Here, s is the maximum height from the point where ball is thrown

So,     s = 55(1.7) - (0.5)(32.2)(1.7)(1.7)

⇒ s = 140 feet

So at a height of 140 feet + 10 feet (initial height) = 150 feet, the ball acts as a freely dropped body.

Here,      u = 0

               a = +32.2

               s= 150

<em>s = ut + (1/2)at²</em>

⇒  150 = 0.5 ( 32.2) (t²)

⇒  t² = 300/32.2 = 9.31

⇒ t = 3.05 sec

So total time = 1.7 + 3.05 = 4.75 seconds

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