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3 years ago

CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION? 4x+2=7x-37

Mathematics

2 answers:

Blababa [14]3 years ago

6 0

Ok

4x+2-2= 7x-37-2

4x= 7x - 35

4x-7x= 7x-7x-35

-3x/-3 = -35/-3

X = 35/3

grin007 [14]3 years ago

5 0

<h2>Answer:</h2><h2>x = 13</h2><h2 /><h2>Hope this helps!!</h2>

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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.

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3 years ago

Write an equation of the line that is parallel to 2x + 4y = 6 and passes through the point (6, 4). A) y = 2x + 4 B) y = 2x - 8 C

Lynna [10]

Answer:

$\large\boxed{D.\ y=-\dfrac{1}{2}x+7}$

Step-by-step explanation:

$\text{Let}\\k:\ A_1x+B_1=C_1\\\\l:\ A_2x+B_2y=C_2\\\\\text{then}\\\\k\ \parallel\ l\iff A_1=A_2\ \wedge\ B_1=B_2\\==========================$

$\text{We have the equation:}\ 2x+4y=6\\\\\text{Therefore the equation of a parallel line is:}\ 2x+4y=C\\\\\text{Put the coordinates of the given point to the equation and solve for}\ C:\\\\(6,\ 4)\to x=6,\ y=4\\\\C=2(6)+4(4)\\C=12+16\\C=28\\\\\text{The equation in the standard form:}\\\\2x+4y=28$

$\text{Convert to the slope-intercept form}\ y=mx+b:\\\\2x+4y=28\qquad\text{subtract}\ 2x\ \text{from both sides}\\\\4y=-2x+28\qquad\text{divide both sides by 4}\\\\y=-\dfrac{2}{4}x+\dfrac{28}{4}\\\\y=-\dfrac{1}{2}x+7$

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4 years ago

2k - 2j when j=4&k=6 replace the variable with the given amount and solve the problem

eimsori [14]

Answer:

Step-by-step explanation:

first, plug in 6 for k. (it doesn't matter which value you plug in first; you can start with j or k, but i'm just starting with k since that is the variable that comes first in the expression)

2k - 2j => 2(6) - 2j

then plug in 4 for j.

2(6) - 2j => 2(6) - 2(4)

now your expression is 2(6) - 2(4). simplify by multiplying 2(6) and then -2(4) (again, it doesn't matter which one you multiply first, as long as you make sure that the minus sign between 2(6) and 2(4) gets attached to the 2 in 2(4).)

2(6) - 2(4) => 12 - 2(4)

12 - 2(4) => 12 - 8

subtract 8 from 12.

12 - 8 = 4

your answer is 4, therefore when j = 4 and k = 6, 2k - 2j = 4.

i hope this helps! have a nice day <3

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3 years ago

What is 2/6 times 3 in its simplest form

lara [203]

Answer:

Step-by-step explanation:

$\frac{2}{6} \times 3 = \frac{2}{6} \times \frac{3}{1}$

$\frac{2}{6} \times \frac{3}{1} = \frac{6}{6} = 1$

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devlian [24]

Answer:

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Step-by-step explanation:

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