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3 years ago

Which model represents 1/2 divided by 1/6?

Mathematics

1 answer:

kvasek [131]3 years ago

3 0

Answer:

The answer I believe is : D.

~ Hope this is correct, if I'm wrong I'm sorry, have a gr8 day/ night my friend!~

Step-by-step explanation:

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Change 2x+y=-4 into slope-intercept form with steps

nikitadnepr [17]

Answer : y = -2x - 4

All you have to do is put y on one side and the 2x on the other side, just make sure to make it negative since you are subtracting it to go on the other side

8 0

3 years ago

You are situated 300 feet from the base of Tower Glitz Plaza watching an external elevator descend down the side of the building

nordsb [41]

Answer:

18.66 ft/s

Step-by-step explanation:

The distance between you and the elevator is given by:

$h=\sqrt{x^2+y^2}$

The rate of change for the distance between you and the elevator is given by:

$\frac{dh}{dt}=\frac{dh}{dy}*\frac{dy}{dt}$

$-16=\frac{dh}{dy}*\frac{dy}{dt}$

$\frac{dh}{dy}=\frac{d}{dy} (\sqrt{x^2+y^2})\\$

Applying the chain rule:

$u=x^2+y^2\\\frac{dh}{dy}=\frac{d\sqrt u}{du} *\frac{du}{dy}\\\frac{dh}{dy}=\frac{1}{2\sqrt u} *2y\\\frac{dh}{dy}=\frac{y}{\sqrt {(x^2+y^2)}}$

Therefore, at x=300 and y = 500, dy/dt is:

$-16=\frac{y}{\sqrt {(x^2+y^2)}}*\frac{dy}{dt}\\-16=\frac{500}{\sqrt {(300^2+500^2)}}*\frac{dy}{dt}\\\frac{dy}{dt}=-18.66\ ft/s$

The elevator is descending at 18.66 ft/s.

7 0

3 years ago

What is the solution set for the open sentence with the given replacement set y2 -3=y+ [0,1,3,4]

stich3 [128]

It is three cuz when replace with three it is all equal:
3(2)-3=3
6-3=3
3=3
please rate as brainliest and hope this helps :)

7 0

3 years ago

. 2. Find the volume of the right cone. 4 cm 3 cm

Ainat [17]

Answer:

i dont get it please alaborate

Step-by-step explanation:

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 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:

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

2 years ago