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3 years ago

Factor the expression completely. xy^4-y^2

Mathematics

2 answers:

Vilka [71]3 years ago

7 0

Answer: (xy^2 + y)(xy^2 - y)

Step-by-step explanation:

sattari [20]3 years ago

3 0

Answer:

$y^2(xy^2-1)$

Step-by-step explanation:

1) look at the two terms and see if they have a GCF

*both terms have a gcf of y^2*

y^2(xy^2-1)

Hope this helps!

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What do you go by for the pattern 360,60,10 would it be add 60, divided by 6 ,multiply by 6 or subtract 300?

Elina [12.6K]

Answer:

divided by 6

Step-by-step explanation:

Given pattern

360,60,10

would it be add 60

lets

add 60 to each term

360 +60 = 420

but next term is 60, hence it incorrect choice

divided by 6

lets divide each term by 6

360/6 = 60 which is the next term in the series as well

60/6 = 10 which is also the next term in the series as well

hence divided by 6 is the correct option.

multiply by 6

multiply by 6 to each term

360 *60 = 21600

but next term is 60, hence it incorrect choice

subtract 60 from each term

360 -300 = 60 which is the next term in the series

60 -300 = -240 which is not same the next term in the series that is 10

hence this is incorrect choice

8 0

3 years ago

The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6

IrinaK [193]

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$ , in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$ .

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$ .

This is P(X = 4). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0

3 years ago

Can someone help me with this math homework please!

sladkih [1.3K]

Answer:

1st option.

0 to 100 miles per hour

Step-by-step explanation:

as we can see the question states that a particular car' s gas mileage DEPENDS upon its speed.

Since, the independent variable is the domain of the function the speed of the car will act as the domain of the given function.

out of all the option option 1 gives us values of speed( cause its unit is miles/ hour - unit of speed).

so the domain is

0 to 100 miles per hour.

7 0

3 years ago

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djyliett [7]

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3 years ago

At 8:00 A.M., the temperature is 53 °F. At 11:00 A.M., it is 69 °F.

NARA [144]

The average change per hour is B 40.7

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