Which of these are like terms?

Covert 5 miles to meters round to the nearest meter

velikii [3]

8046.72~~~~~8047 that’s the answer

8 0

3 years ago

Determine the miles traveled in 5 hours at a rate of 189 miles in 3 hours

Salsk061 [2.6K]

Hey there,

Question : <span>Determine the miles traveled in 5 hours at a rate of 189 miles in 3 hours

Answer :
3 hours = 189 miles
1 hour = 189 miles / 3 hours
= 63 miles
5 hours = 63 miles x 5 hours
= 315 miles

Hope this helps :))

<em>~Top♥</em>
</span>

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

Please solve!!! <br><br> In a hurry!

boyakko [2]

Answer:

The answer is <u>y=5</u>

Step-by-step explanation:

3y=15

you divide 3 into 3 and 15

3y/3=15/3

y=5

3 0

3 years ago

The equation represents a hyperbola centered at the origin with a directrix of . what is the value of b? 10 16 20 26

Rom4ik [11]

the value of b is = 10.

The equation of a hyperbola is x^2/24^2 - y^2/ (10)^2= 1.

<h3>What is hyperbola?</h3>

The geometric characteristics of a hyperbola or the equations for which it is the solution set characterize it as a particular kind of smooth curve that lies in a plane. Mirror reflections of each other that resemble two infinite bows make up a hyperbola's two connected components or branches.

<h3>What is the general formula for hyperbola?</h3>

The general formula for hyperbola = (x - h)²/a²- (y - k)²/(b)² = 1

<h3>According to the given information:</h3>

x²/24 - y²/(b)² = 1

(x - 0)²/24 - (y -0)²/(b)² = 1

a=24,h=0 and k=0

Now equation of the directrix

x=a²/c...(1)

and we know x=576/26...(2)

Therefore from 1 and 2 we get

24²/c=576/26.

isolate the c so we get,

C=26

C= center of focii

c = √(a² + b³)

c² = a² + b²

b = c² - a²

b = 10

So we get the value of b is 10.

Therefore the equation of a hyperbola is x²/24² - y²/ (10)² = 1.

To know more about Hyperbola visit:

brainly.com/question/12919612

#SPJ4

8 0

2 years ago