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Kobotan [32]
2 years ago
7

13069 rounded to the nearest thousand

Mathematics
1 answer:
Alexxandr [17]2 years ago
7 0
The anwser is 13000 all you do is look at the three and the three is bigger than the zero. so it goies to 13000
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Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and
Olenka [21]

Answer:

The probability that the series lasts exactly four games is 3p(1-p)(p^{2} + (1 - p)^{2})

Step-by-step explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

P(A \cap B) = P(A)*P(B)

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(A) = P(X = 2) = C_{3,2}.p^{2}.(1-p)^{1} = 3p^{2}(1-p)

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

P(A4) = P(A)*P(B) = 3p^{2}(1-p)*p = 3p^{3}(1-p)

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(A) = P(X = 1) = C_{3,1}.p^{1}.(1-p)^{2} = 3p(1-p)^{2}

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

P(B4) = P(A)*P(B) = 3p(1-p)^{2}*(1-p) = 3p(1-p)^{3}

Probability that the series lasts exactly four games:

p = P(A4) + P(B4) = 3p^{3}(1-p) + 3p(1-p)^{3} = 3p(1-p)(p^{2} + (1 - p)^{2})

The probability that the series lasts exactly four games is 3p(1-p)(p^{2} + (1 - p)^{2})

8 0
3 years ago
Nora tracked variations in temperature for five days to show how much above or below average the actual temperature was.
stepladder [879]

Answer:

Wednesday

Step-by-step explanation:

1. First find the position of each variable on the number line.

2. The average temperature is 0 because it is in the center.

3. The answer cannot be Monday because it is above so it is Wednesday.

5 0
3 years ago
Read 2 more answers
The product of (a − b)(a − b) is a perfect square trinomial
SOVA2 [1]

Answer:

always

Step-by-step explanation:

this product can also be written as a^2 - ab - ab + b^2

which is a^2 - 2(ab) + b^2

a perfect square trinomials equation is

a^2 + 2(ab) + b^2

and this qualifies

6 0
2 years ago
Cual es la rais cuadrada de 2
sergejj [24]
La raíz cuadrada de 2 es 1.414
3 0
3 years ago
Which equation represents the circle described? The radius is 2 units. The center is the same as the center of a circle whose eq
Serhud [2]

Answer:

(x-4)^2+(y-3)^2=2^2

Step-by-step explanation:

The given circle has equation; x^2+y^2-8x-6y+24=0

Comparing to the general equation of the circle:  x^2+y^2+2ax+2by+c=0

We have 2a=-8\implies a=-4 and 2b=-6\implies b=-3

The center of this circle is (-a,-b)=(4,3).

The required circle has radius r=2 units.

The equation of a circle, given the center (h,k) and radius r, is given by:

(x-h)^2+(y-k)^2=r^2

We substitute the values to obtain (x-4)^2+(y-3)^2=2^2

6 0
3 years ago
Read 2 more answers
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