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3 years ago

Michael has two options to invest $10,000. Which option will earn the most money?

Mathematics

2 answers:

irina [24]3 years ago

7 0

Answer:

C. is correct

Step-by-step explanation:

10,000*0.06=600*10=6,000

10,000*0.05=500*12=6,000

So, they are both the same :[

Art [367]3 years ago

3 0

They are both the same because when you Multiply 10,000 and 6% you get 600 and then you multiply that by 10 years you get 6,000 dollars and than you do the same with the 5% but multiply the sum by 12 years and you get 6,000 also so gives you that they are both the some income so I would say Michael should just keep the $10,000

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Use substitution to solve system of equation<br>x-2y=12<br>Y = 3x + 14

andrew11 [14]

The answer is x=-8 and y=-10

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3 years ago

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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

Volume with unit cubes 2

ololo11 [35]

Answer:

Volume = length * width * hight

4 0

4 years ago

Statistics Probability

andriy [413]

Answer:

Explained below.

Step-by-step explanation:

Denote the variable as follows:

M = male student

F = female student

Y = ate breakfast

N = did not ate breakfast

(a)

Compute the probability that a randomly selected student ate breakfast as follows:

$P(Y)=\frac{n(Y)}{N}\\\\=\frac{198}{330}\\\\=0.60$

(b)

Compute the probability that a randomly selected student is female and ate breakfast as follows:

$P(F\cap Y)=\frac{n(F\cap Y)}{N}\\\\=\frac{121}{330}\\\\=0.367$

(c)

Compute the probability a randomly selected student is male, given that the student ate breakfast as follows:

$P(M|Y)=\frac{n(M\cap Y)}{n(Y)}\\\\=\frac{77}{198}\\\\=0.389$

(d)

Compute the probability that a randomly selected student ate breakfast, given that the student is male as follows:

$P(Y|M)=\frac{n(Y\cap M)}{n(M)}\\\\=\frac{77}{154}\\\\=0.50$

(e)

Compute probability of the student selected "is male" or "did not eat breakfast" as follows:

$P(M\cup N)=P(M)+P(N)-P(M\cap N)\\\\=\frac{n(M)}{N}+\frac{n(N)}{N}-\frac{n(M\cap N)}{N}\\\\=\frac{n(M)-n(N)-n(M\cap N)}{N}\\\\=\frac{154+132-77}{330}\\\\=0.633$

(f)

Compute the probability of "is male and did not eat breakfast as follows:

$P(M\cap N)=\frac{n(M\cap N)}{N}\\\\=\frac{77}{330}\\\\=0.233$

4 0

3 years ago

If LM = 22 and MN = 15, find LN

levacccp [35]

Assuming that point M lies on line segment LN, then by the segment addition postulate, we can say

LN = LM + MN
LN = 22 + 15
LN = 37

8 0

4 years ago

Read 2 more answers