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IRINA_888 [86]
3 years ago
10

How manygallons of gas will I use to travel100 miles? 300 miles?​

Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer:

8 gallons to travel 100 miles. 24 gallons for 300 miles

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The volume of the composite figure is 140 in³

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Marina's New Year's resolution is to ride her bike 5000 miles.
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m5000=d

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or d5000=m

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How many solutions does the equation 22 = 81 have?
agasfer [191]

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No solutions.

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Solve: −3(5+8x)−20≤−11
Tatiana [17]
-3(5 + 8x) - 20 ≤ -11  |use distributive property: a(b + c) = ab + ac

-15 - 24x - 20 ≤ -11

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3 years ago
Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How
HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Parameter of 5.2 per square yard:

This means that \mu = 5.2r, in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

P(X \geq 1) = 1 - P(X = 0) = 0.99

Thus:

P(X = 0) = 1 - 0.99 = 0.01

We have that:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}

Then

e^{-5.2r} = 0.01

\ln{e^{-5.2r}} = \ln{0.01}

-5.2r = \ln{0.01}

r = -\frac{\ln{0.01}}{5.2}

r = 0.89

Thus, the radius should be of at least 0.89.

Another example of a Poisson distribution is found at brainly.com/question/24098004

3 0
3 years ago
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