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4 years ago

Please answer this correctly

Mathematics

1 answer:

max2010maxim [7]4 years ago

3 0

Answer:

Yellow model, 20%

Step-by-step explanation:

There are 25 rooms that are vacant, out of 125. This means that 20% of the rooms are vacant. The yellow model displays this correctly because the sections are 1/5 or 20% out of 125.

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BRAINLIEST!!!<br> 12. Determine whether each pair of functions are inverse functions.

denpristay [2]

Answer:

The answer is 2nd point

Step-by-step explanation:

In this question, inverse of f(x) is equals to g(x). First, you must know how to find the inverse of f(x). In order to find the g(x), you must let f(x) = y and make x thd subject :

Let y = (10x+9)/8

y = (10x+9)/8

8y = 10x + 9

8y - 9 = 10x

(8y-9)/10 = x

x = (8y-9)/10

g(x) = (8x-9)/10 (Remember to convert y back to x)

Let y = x - 9

y = x - 9

y + 9 = x

x = y + 9

g(x) = x + 3

Next, by looking in no.1, the inverse of f(x) is wrong and in no.2, it shows the correct form of inverse. So only 2 is an inverse function

4 0

3 years ago

A quadratic function and an exponential function are graphed below. How do the decay rates of the functions compareover the inte

Kobotan [32]

To check the decay rate, we need to check the variation in y-axis.

Since our interval is

$-2We need to evaluate both function at those limits.At x = -2, we have a value of 4 for both of them, at x = 0 we have 1 for the exponential function and 0 to the quadratic function. Let's call the exponential f(x), and the quadratic g(x).[tex]\begin{gathered} f(-2)=g(-2)=4 \\ f(0)=1 \\ g(0)=0 \end{gathered}$

To compare the decay rates we need to check the variation on the y-axis of both functions.

$\begin{gathered} \Delta y_1=f(-2)-f(0)=4-1=3 \\ \Delta y_2=g(-2)-g(0)=4-0=4 \end{gathered}$

Now, we calculate their ratio to find how they compare:

$\frac{\Delta y_1}{\Delta y_2}=\frac{3}{4}$

This tell us that the exponential function decays at three-fourths the rate of the quadratic function.

And this is the fourth option.

4 0

1 year 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

Let n = 60! + 55! + 50! the unit digit of n and a number of digits to the left of the units digits are consecutive zeros before

Archy [21]

n=8320987125437793528348710310901843271933297732006335848276356027645952000000000000

So you can see there are 12 zeros before you hit the first non-zero, which is 2.

8 0

3 years ago

Write a formula for a line that passes through the point (x,y) = (-3,5) and has an undefinedslopes.

Vladimir [108]

A line with undefined slope has an equation x=a.

So, for a line that passes through the point (x,y) = (-3,5) ,

has the equation x= - 3.

5 0

4 years ago