Below are two images with scale copies of each other. Determine the scale factor for each of the following:

Can you guys help me solve this ?

Serhud [2]

Answer:

10i

Step-by-step explanation:

4 0

3 years ago

Select all pairs of corresponding angles. Assume the lines are parallel.

Elena-2011 [213]

Step-by-step explanation:

Following are the pair of corresponding angles.

1 and 5

2 and 6

3 and 7

4 and 8

3 0

3 years ago

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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

Solve for n <br> X^n*x^3=x^13

kykrilka [37]

x^n*x^3=x^13

x^(n+3)=x^13

n+3=13

n=10

5 0

3 years ago

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Please help me it says find the measure of c and b and I don’t know how if you can help me with this exam it would be much appre

noname [10]

Answer:

b=20

C=160

Step-by-step explanation:

160+b= 180 (sum of angle on the straight line)

b=180-160

b=20

b+c =180 (sum of angle on the straight line)

20+c=180

c=180-20

C=160

6 0

2 years ago