Carla use two and three cups and one and three eight how much does she use in all

Alex_Xolod [135]

87 is greater of course

6 0

3 years ago

A cell phone tower is anchored by two cables on each side for support. The cables stretch from the top of the tower to the groun

kramer

Answer:

659.6 ft

Step-by-step explanation:

x = 140 / tan 23

x = 329.82(rounded to hundredth)

so total ground distance between the cables = 2x = 2(329.82) = 659.64 feet

4 0

3 years ago

Evaluate the function.

RSB [31]

It would be d
because -5•-2=10
and 10+3=13

4 0

3 years ago

Read 2 more answers

1. What is the average rate of change of the function f(x)=5(2)^x from x = 1 to x = 5?

Gnom [1K]

Answer:

The last one is B.

Step-by-step explanation:

4 0

3 years ago

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