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3 years ago

Solve for the variable in 6/18 = X/36. O A 9 OB) 6 O C) 12 OD) 3

Mathematics

2 answers:

zimovet [89]3 years ago

6 0

Answer:

Step-by-step explanation:

6/18 = X/36

or, 6 multiply by 36 = x multiply by 18

or, 216 = 18x

or, 216/ 18 = x

or, x= 12

MAVERICK [17]3 years ago

4 0

Answer:

Step-by-step explanation:

6/18 can be simplified to 3/9 which can then be simplified to 1/3

12/36 can be simplified to 6/18 which can be simplified to 3/9 which can then be simplified to 1/3 making 12 the correct answer

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How to now which one is bigger 16.45 or 16.454

artcher [175]

16.454 because there is an extra 4 after the 5

7 0

3 years ago

Read 2 more answers

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

Write a sine and cosine function that models the data in the table. I need steps to both for a, b, c, and d.

andrezito [222]

Answer(s):

$\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2} \\ y = -29cos\:\frac{\pi}{6}x + 44\frac{1}{2}$

Step-by-step explanation:

$\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-3} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{6}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29$

OR

$\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29$

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the centre photograph displays the trigonometric graph of $\displaystyle y = -29sin\:\frac{\pi}{6}x + 44\frac{1}{2},$ in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [centre photograph] is shifted $\displaystyle 3\:units$ to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD $\displaystyle 3\:units,$ which means the C-term will be negative, and by perfourming your calculations, you will arrive at $\displaystyle \boxed{3} = \frac{-\frac{\pi}{2}}{\frac{\pi}{6}}.$ So, the sine graph of the cosine graph, accourding to the horisontal shift, is $\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2}.$ Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit $\displaystyle [12, 15\frac{1}{2}],$ from there to the y-intercept of $\displaystyle [0, 15\frac{1}{2}],$ they are obviously $\displaystyle 12\:units$ apart, telling you that the period of the graph is $\displaystyle 12.$ Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at $\displaystyle y = 44\frac{1}{2},$ in which each crest is extended twenty-nine units beyond the midline, hence, your amplitude. Now, there is one more piese of information you should know -- the cosine graph in the photograph farthest to the right is the OPPOCITE of the cosine graph in the photograph farthest to the left, and the reason for this is because of the negative inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of any trigonometric equation, the whole graph reflects over the midline. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

3 0

2 years ago

a student claims that the equation y=7 is not a linear equation because it does not have the from y=m x+b.do you agree or disagr

KIM [24]

That is false.
y=7 is just a simplified way of writing y = 0 x 1 + 7

7 0

3 years ago

The U.S. Department of Agriculture records data on farm acreage and number of farms by county for every county in the country. T

lisabon 2012 [21]

Solution :

From the table,

Number of countries = $N_i$

Mean = $\overline Y_i$

Standard deviation = $\sigma_i$

Margin of error of B = 50,000 acres

So we determine the $\text{sample size n}$ and then allocate the four regions of $n_1, n_2, n_3$ and $n_4$ .

Under the $\text{proportional allocation}$ , the $\text{equation}$ for the value of $n_1$ which yields $V(\overline Y_{st}) = D = \frac{B^2}{H}$ is given by :