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3 years ago

What is a value of y that satisfies the inequality below? y3≤12

Mathematics

2 answers:

STALIN [3.7K]3 years ago

8 0

Answer:

Step-by-step explanation:

because 4 Times 3 equals 12 so less than or equal to hope this helps

Serga [27]3 years ago

8 0

Answer:

y =4

Step-by-step explanation:

3y=12

3y/3=12/3

y=4

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For each ordered pair, determine whether it is a solution to the system of equations.

-Dominant- [34]

The only ordered pair that is a solution to the given system of equations is (-2, -6)

<h3>System of Linear Equations </h3>

From the question, we are to determine if each ordered pair is a solution to the given system of equations

The given system of equations is

-9x + 2y = 6

5x - 3y = 8

For (7, 9)

That is,

x = 7, y = 9

Putting the values into the first equation

Is -9(7) + 2(9) = 6

-63 + 18 = 6

-45 ≠ 6

Thus, (7,9) is not a solution

For (0, 3)

That is,

x = 0, y = 3

Putting the values into the first equation

Is -9(0) + 2(3) = 6

0 + 6 = 6

6 = 6

The ordered pair satisfies the first equation

Testing for the second equation

Is 5(0) - 3(3) = 8

0 - 9 = 8

-9 ≠ 8

Thus, (0, 3) is not a solution

For (5, -4)

That is,

x = 5, y = -4

Putting the values into the first equation

Is -9(5) + 2(-4) = 6

-45 - 8 = 6

-53 ≠ 6

Thus, (-5,4) is not a solution

For (-2, -6)

That is,

x = -2, y = -6

Putting the values into the first equation

Is -9(-2) + 2(-6) = 6

18 - 12 = 6

6 = 6

The ordered pair satisfies the first equation

Testing for the second equation

Is 5(-2) -3(-6) = 8

-10 + 18 = 8

8 = 8

The ordered pair satisfies the second equation

∴ The ordered pair that is a solution to the system of equations is (-2, -6)

Hence, the only ordered pair that is a solution to the given system of equations is (-2, -6)

Learn more on System of equations here: brainly.com/question/8630769

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1 year ago

William is drafting his fantasy basketball team. He needs to select one player for each position. The following table shows how

denpristay [2]

The corrrect answer is 77

4 0

3 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: