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2 years ago

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1 answer:

8 0

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{4}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{2}{4}\implies \cfrac{1}{2}$

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I am greater than 15 I am less than 21 I have 8 ones

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

bc 8 digits in the ones place, and less than 21 greater than 15

7 0

3 years ago

Read 2 more answers

What is the largest prime number?

Lady bird [3.3K]

Answer:

28258933-1

Step-by-step explanation:

hope it helps you

6 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

Idek what to do, help

stepladder [879]

Distribute the 5 to the (2+y). Parenthisis always come first. after that combine the 5y and the -y. you should end up with 2-(+4y) +2. add the -2 and +2. you should be left with 4y

8 0

3 years ago

A property map is drawn at a scale of 1:1000. Calculate the corresponding actual distance if the distance between two points on

zvonat [6]

Answer:10 m

Step-by-step explanation:

Given

Scaling ratio 1:1000

i.e. one unit in map is equal to 1000 unit in actual distance

therefore if the distance between two points is 1 cm in map then

it is 1000 cm in reality

8 0

3 years ago