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3 years ago

Probabilities. Please help

Mathematics

1 answer:

Anon25 [30]3 years ago

3 0

In throwing a dice, there are 6 possible outcomes, namely 1, 2, 3, 4, 5 and 6.

P=(no. of favourable outcomes)/(no. of possible outcomes)

When we multiply the probability of an event with no. of trials, the we get the expected frequency of that event.

1) P(getting 1)=1/6

expected frequency of getting 1

=P(getting)1)×no. of trials

=1/6×600=100

But it is given that one is scored 200 times, so it makes a large difference from expected frequency, which is 100.

Therefore, the dice is not fair.

2) P(getting a tail)=0.3

estimate for the number of times the coin will land on a tail

=expected frequency of getting a tail

=P(getting a tail)×no. of trials

=0.3×150=45

3)P(getting a six)=2/3

expected frequency of getting a six

=P(getting a six)×no. of trials

=2/3×300=200

4)P(getting a three)=0.5

expected frequency of getting a three

=P(getting a three)×350

=0.5×350=175

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A pair of tow trucks are pulling on a trailer that is stuck

Varvara68 [4.7K]

Answer:

R=10,244 Lb

Step-by-step explanation:

In order to solve this problem, we can start by drawing the situation. (See attached picture).

We can look at the forces applied by the trucks as vectors, so in this case, in order to find the resultant force, we can add the two vectors. You can do so by drawing one force after the other and join the start of the first force with the end of the second force to form a triangle. The missing side will be the resultant force. In that triangle I drew there, the inner angle will be found by subtracting 180°-110°=70°

In this case we can find the resultant force by using the law of cosines:

$A^{2}=B^{2}+C^{2}-2AB cos \alpha$

so we can use the data given by the problem:

$R^{2}=T_{1}^{2}+T_{2}^{2}-2T_{1}T_{2} cos \alpha$

so we can solve this for R, so we get:

$R=\sqrt{T_{1}^{2}+T_{2}^{2}-2T_{1}T_{2} cos \alpha}$

now we can substitute:

$R=\sqrt{(10,000Lb)^{2}+(7,500Lb)^{2}-2(10,000Lb)(7,500Lb) cos (70^{0})}$

which yields:

R=10,244Lb

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3 years ago

A hopper contains 20 ping-pong balls. There are 6 red, 4 white, 3 green, 2 yellow,

Aleks [24]

Answer:

0.079 would be the correct answer

Step-by-step explanation:

you just do 3/10 x 5/19 and then you get the answer

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3 years ago

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin w

Goshia [24]

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability of getting heads is :

$P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})$

$= \dfrac{P( \{Tails, Heads\}) \cap \{Tails, Heads,( Heads ,Tails)\})}{ {P( \{Tails, Heads,( Heads ,Tails)\}}}$

$= \dfrac{P( \{Tails, Heads\}) }{ {P( \{Tails, Heads,( Heads ,Tails)\}}}$

$= \dfrac{P( \{Tails, Heads\}) } { {P( Tails, Heads) +P( Heads ,Tails)}}$

$=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}$

$=\dfrac{(1-p)*p}{2(1-p)*p}$

$=\dfrac{1}{2}$

Thus; the probability of getting heads is $\dfrac{1}{2}$ which typically implies that the coin is fair

(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

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3 years ago

–5x+4<br><br> constant<br> coefficient:<br> variable:

SSSSS [86.1K]

Answer:

constant: 4

coefficient:-5

variable:x

Step-by-step explanation:

You can easily search up the definition of each word and figure it out just saying. It's fine if u wanna do this-

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3 years ago

Please, help me on this question, guys! I know the answer for (a) so I'll write down, too. For (b) I tried "$89,000 because that

Alexeev081 [22]

Let x be a random variable representing the price of a Congo-imported black diamond. Let the higher price be p. Then,
P(x < p) = P(x < (p - mean)/sd) = P(x < (p - 60,430)/21,958.08) = P(z < 2)
Therefore,
(p - 60,430)/21,958.08 = 2
p - 60,430 = 2 x 21,958.08 = 43,916.16
p = 34,916.16 + 60,430 = 104.346.16

Therefore, The required price is $104,346.16

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3 years ago