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2 years ago

What is the answer 4+3x = -2y

1 answer:

6 0

First add 2y to both sides
so 4 + 3x + 2y = 0
then subtract 4 from both sides
so 3x + 2y = -4
that is standard form
now subtract 3x from both sides
2y = 3x - 4
now divide by 2 on both sides
y = 1.5x - 2
This is solved for y and is slope intercept form
starting from original subtract 4 from both sides
so 3x = -2x -4
now divide by 3 on both sides
so x = -2/3x -4/3
this is solved for x
hope any of that is what you needed.

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If the distribution is really (5.43,0.54)

defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

$\mu = 5.43, \sigma = 0.54$

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.79 - 5.43}{0.54}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

7 0

3 years ago

First-person to answer correctly with an explanation get brainiest.

Aneli [31]

Answer:

x = 8

Step-by-step explanation:

I think this is the answer if I am wrong please do not report me

I hope this helps! :}

4 0

2 years ago

Read 2 more answers

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 trials before q successes 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: