Please help me wonderful people! its math!!

In algibratic fractions what is 1/2x-1/3y

Doss [256]

Simplify each term.

1/2X - 1/3Y

Combine 1/2x and −1/3y using a common denominator.

To write −1/3y as a fraction with a common denominator, multiply by 2x/2x.

(1 / 2x )(3y / 3y) + (−1 / 3y) (2x / 2x)

Write each expression with a common denominator of 6xy, by
multiplying each by an appropriate factor of 1.

1 (3y)/6xy − 1 (2x)/6xy

Combine the numerators over the common denominator.

1(3y) − (1(2x))/6xy

Simplify the numerator.

3y−(1(2x))/6xy

Simplify.

3y−(2x)/6xy

3y−2x/6xy

Simplify with factoring out.

Reorder 3y and −2x.

−2x+3y/6xy

Factor −1 out of −2x.

−1(2x)+3y/6xy

Factor −1 out of 3y.

−1(2x)−1(−3y)/6xy

Factor −1 out of −1(2x)−1(−3y).

−1(2x−3y)/6xy

Move the negative in front of the fraction.

−2x−3y/6xy

5 0

3 years ago

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Jasmine wants to build a 2 2/5 meter long garden path paved with square stones that measure 1/4 meter on each side. There will b

Allisa [31]

The answer is 9 3/5

3 0

4 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

worty [1.4K]

Answer:

a) $P(X \leq 100) = 1- e^{-0.01342*100} =0.7387$

$P(X \leq 200) = 1- e^{-0.01342*200} =0.9317$

$P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930$

b) $P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498$

c) $m = \frac{ln(0.5)}{-0.01342}=51.65$

d) $a = \frac{ln(0.05)}{-0.01342}=223.23$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

Solution to the problem

For this case we have that X is represented by the following distribution:

$X\sim Exp (\lambda=0.01342)$

Is important to remember that th cumulative distribution for X is given by:

$F(X) =P(X \leq x) = 1-e^{-\lambda x}$

Part a

For this case we want this probability:

$P(X \leq 100)$

And using the cumulative distribution function we have this:

$P(X \leq 100) = 1- e^{-0.01342*100} =0.7387$

$P(X \leq 200) = 1- e^{-0.01342*200} =0.9317$

$P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930$

Part b

Since we want the probability that the man exceeds the mean by more than 2 deviations

For this case the mean is given by:

$\mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516$

And by properties the deviation is the same value $\sigma = 74.516$

So then 2 deviations correspond to 2*74.516=149.03

And the want this probability:

$P(X > 74.516+149.03) = P(X>223.547)$

And we can find this probability using the complement rule:

$P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498$

Part c

For the median we need to find a value of m such that:

$P(X \leq m) = 0.5$

If we use the cumulative distribution function we got:

$1-e^{-0.01342 m} =0.5$

And if we solve for m we got this:

$0.5 = e^{-0.01342 m}$

If we apply natural log on both sides we got:

$ln(0.5) = -0.01342 m$

$m = \frac{ln(0.5)}{-0.01342}=51.65$

Part d

For this case we have this equation:

$P(X\leq a) = 0.95$

If we apply the cumulative distribution function we got:

$1-e^{-0.01342*a} =0.95$

If w solve for a we can do this:

$0.05= e^{-0.01342 a}$

Using natural log on btoh sides we got:

$ln(0.05) = -0.01342 a$

$a = \frac{ln(0.05)}{-0.01342}=223.23$

5 0

3 years ago

6y is less than 18 how do i do it

tia_tia [17]

So if the equation is 6y<18, you have to get the y alone so you have to do inverse operations.
Since 6 is being multiplied by y, you have to divide the whole equation by 6. You are left with y<3, which is the answer. Do you understand?

7 0

3 years ago

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Simplify 1/3n (-6+27m-51p)

marishachu [46]

1/3n * -6 = -2n
1/3n * 27m = 9nm
1/3n* -51p = -17np
-2n + 9nm - 17np

7 0

3 years ago

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