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Nitella [24]
3 years ago
5

Which is larger 90ft or 32yd?

Mathematics
2 answers:
hichkok12 [17]3 years ago
3 0
32yd is longer as 32yd is equivalent to 96ft
ale4655 [162]3 years ago
3 0
It would be 32 yards.
32 yards = 96 foot. 
90 foot = 30 yards.

Hope This Helps You!
Good Luck Studying :)


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Solve the proportion for x. 4/(5x) = 8/9
Verdich [7]
4     8
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5x    9

5x*2=9
5x=4.5
x=4.5/5 or 0.9
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How do I solve this equation q(a+y)=67y+93
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A= 67y+93/q-y; q=/0
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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

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Answer: I believe the answer is 9 days.

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4 0
4 years ago
Read 2 more answers
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Tatiana [17]
Assuming that you’re meant to distribute (you didn’t list expressions), this would be your answer:

-12x + 20

Hope this helps!
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