1. I need all these answers<br><br> ?

There are five sales agents in a certain real estate office. One month Andy sold twice as many properties as Ellen, Bob sold 3 m

Novosadov [1.4K]

Answer:

Carry

Step-by-step explanation:

Here let’s take any arbitrary value for the number of properties sold by Ellen.

Let the number of properties sold by Ellen he 10. We were told Andy sold twice of this, this means he sold 20 properties

Bob sold 3 more than Ellen, meaning he sold 13.

Carry sold twice of Bob meaning he sold 26

Dora sold the addition of Bob and Ellen = 13 + 10 = 23

Carry sold 26 and this makes him the highest seller.

7 0

4 years ago

Open the picture, help me and get 10 points

Hitman42 [59]

13 feet wide.
156/12=13
Brainliest plz

4 0

3 years ago

A point 4 units to the left of the y-axis is reflected across the y-axis. What is the distance between the point and its reflect

prisoha [69]

8 units, what tends to help is to visualize or draw what the problem is asking of you.

5 0

3 years ago

Read 2 more answers

Population density is a measure of how crowded an area is. The population density of Australia is approximately 7.6 people per s

nasty-shy [4]

Answer:

2957.3 people per square mile

Step-by-step explanation:

2949.7 more than 7.6 is ...

7.6 +2949.7 = 2957.3 . . . . people per square mile in Bangladesh

4 0

3 years ago

Read 2 more answers

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

4 years ago

1. I need all these answers ?