Segment AB is a line segment on a number line. Endpoint A is at

Question 3 of 5

gladu [14]

Answer:

triangle A

Step-by-step explanation:

the area is 16cm²

8×4/2 = 32/2 = 16

5 0

2 years ago

I’m stuck on this please

trasher [3.6K]

Answer:

weak negative

Step-by-step explanation:

because its in negative and its in the thousands place.

3 0

2 years ago

The population of Centerville increases each year. The function C(t) = P(1 +r)^t represents the population of centerville at yea

strojnjashka [21]

Answer:

P represents the population in year 0 ⇒ D

Step-by-step explanation:

* Lets explain the exponential growth function

- The exponential growth function is f(x) = a (1 + r)^t, where a is the initial

amount (at t = 0), (1 + r) is the factor of growth , r is the rate of growth

in decimal ant is the time of growth

* Lets solve the problem

∵ The function C(t) = P(1 + r)^t represents the population of

centerville at year t, where P is the initial population and r is the

rate of increase

- Ex: If your investment is increased 10% annually, then that means

each year, your total has multiplied itself by 110% (the growth factor

is 1 + 10/100 = 1.1)

∴ (1 + r) is the factor grows each year

∵ C(t) = P(1 + r)^t

∴ C depends on P(starting population) , r(the increasing rate and

t(the time in year)

∵ r is the rate of increase means the percentage of increasing , then

0 < r < 1

∴ r is not less than 0

∵ P is the initial amount when t = 0

∴ P represents the population in year 0

4 0

3 years ago

Read 2 more answers

An eperiment conducted found that is the last 25 days a blue pen 5 times. There are 6 blue pens and 14 black pens. What is the e

Bad White [126]

6 out of 14 i think try it

3 0

2 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

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

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

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

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

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

One person, thus $q = 1$ .

The expected value is:

$E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4$

The variance is:

$V = \frac{0.8}{0.04} = 20$

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0

1 year ago

Segment AB is a line segment on a number line. Endpoint A is at –18, and the midpoint is at -6.