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2 years ago

3y^4+9y^3-120y^2=0 Need Answer ASAP

Mathematics

2 answers:

Shkiper50 [21]2 years ago

7 0

Answer:

y = 0 , 5 , − 8

(*￣3￣)╭

andrey2020 [161]2 years ago

5 0

$3y^4+9y^3-120y^2=0\\y=0,5,8$

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9² • (1+4) + 7 – 3 • 4 =

Marizza181 [45]

The answer is your looking for is 400

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2 years ago

Read 2 more answers

You asked 50 randomly chosen employees of a company how many books they read each month. the diagram show the results. there 600

podryga [215]

Answer:

About 432 people read at least one book per month.

Step-by-step explanation:

36/50 people read at least one book, multiply 36/50 by 12 to make the fraction 432/600.

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2 years ago

An airplane is preparing to land at an airport. It is 50400 feet above the ground and is descending at the rate of 3000 feet per

Dovator [93]

Answer:

let t = no of minutes when they are at the same altitude

52200 - 3300t = 2500t

52200 = 3300t + 2500t

52200 = 5800t

t = 52200/5800

t = 9 minutes they will be at the same altitude

The altitude:

9*2500 = 22500 ft

Check on other train

52200 - 9(3300) =

52200 - 29700 = 22500 ft

3 0

1 year ago

Please help!

Reika [66]

The greatest number of friends who can receive a necklace is 6. The reason why is because 6 is the greatest common factor of 72 and 42.

Each friend will have 12 blue beads and 7 red beads. If you divide 6 (GCF) with both numbers you get 12 and 7.

Hope this helps :3

4 0

3 years ago

A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get

julsineya [31]

I have an expression

$\sigma = \sqrt{p(1-p)/n}$

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

$E( (b-p)^2 ) = (-p)^2(1-p) + (1-p)^2p = p(1-p)$

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

$\sigma = \sqrt{p(1-p)/n}$

Plugging in the numbers,

$\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%$

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

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3 years ago