All categories

3 years ago

Elizabeth wants to use a standard number cube to do a simulation for a scenario that

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

3 0

Answer:

Standard numbered cube outcomes are {1, 2, 3, 4, 5, 6}. That makes 6 outcomes, all of them with the same probability to happen. Given that the scenario involves three outcomes with the same probability, one option is to take 1 & 2, 3 & 4 and 5 & 6 as the same result.

You might be interested in

4. Find the length of side EG.

olchik [2.2K]

Answer:

42 cm

Step-by-step explanation:

the first triangle is 7 times bigger than the second triangle. ( 4 x 7 = 28)

so to find out line EG, you times 6 cm by 7. which is 42

3 0

3 years ago

Mariya was asked to solve StartFraction a over negative 13 EndFraction less-than-or-equal-to negative 16 and then graph the solu

lozanna [386]

Answer:

the answer is step 3 because the circle on the number line is open.

pls make mine as the Brainliest answer that would really help

8 0

4 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

Can someone please help!!

Fittoniya [83]

Answer:

20 is the Answer i think

Step-by-step explanation:

6 0

3 years ago

a man who is 6 feet tall casts a shadow that is 11 feet long. At the same time, a tree casts a shadow that is 33 feet long. What

AveGali [126]

Something that is 6ft tall with an 11ft shadow is the same relation as something that is x ft tall with a 33ft shadow. You can set up a proportion and then cross-multiply to solve for x (which is the height of the tree)

7 0

3 years ago