Help me pls :( idk what to do

I need the answer and why

Lorico [155]

Answer:

Brien is incorrect because he skipped step 2

Step-by-step explanation:

4 0

3 years ago

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For f(x)=log(the 2 is below log)2(x), between which two integers does f(40) lie?

Veronika [31]

f(x) = log2 x

f(40) = log2 40

40 = 2^y

2^5 = 32 and 2^6 = 64

so f(40) lies between integers 5 and 6.

4 0

3 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

Mrs. Robertson is preparing manipulatives for her math class. Each student will receive a plastic cup with 20 two-color counters

lana [24]

Answer:

Step-by-step explanation:

I just got this correct on the test

3 0

3 years ago

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Kedar is comparing the costs of phone plans. For phone plan A, the cost is $15.00 to connect and then $0.02 per minute. For phon

zysi [14]

Answer:

The answer is (D): 2 h 52 min

Step-by-step explanation:

write these equations separately as functions:

A(x)=15+0.02m

B(x)=4.69+0.08m

Since we want to find when they are equal, set them equal to each other:

15+0.02m=4.69+0.08m

simplifying this equation gives

10.31+0.02m=0.08m

subtract 0.02m from both sides

10.31=0.06m

divide both sides by 0.06

m=171.833 etc approximately equals 172 minutes

172 min/60=

2 hours and 52 min

Hope this helps!

4 0

4 years ago

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