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Deffense [45]
3 years ago
6

Eric and Frank want to equally share 4/3 feet of rope. what length of rope should each friend get? explain how to use a drawing

to help solve the problem.
Mathematics
1 answer:
blondinia [14]3 years ago
4 0
The rope is 1 and 1/3 long,1 is the same as saying 3/3,
this means that the rope is 3/3 + 1/3 long,so it is 4/3 foot long,
you now want to get half of this number so that each person has an equal length of rope,so multiply the length of rope by a half,
4/3 x 1/2 = 4/6 , which if simplified = 2/3, this means that each friend gets 2/3 foot of rope.

You can draw this in a diagram by drawing the 1 foot of rope, and then dividing it up into 3 sections. Then add the 1/3 foot of rope into the drawing. It should then be easy to see that altogether there are 4 separate sections of rope and that if each friend has two sections then they will be sharing equal amounts, so consequently each friend has two of the thirds of rope, 2/3

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Nikolay [14]

Answer:

What is the question

Step-by-step explanation:

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3 years ago
The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,
Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

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 time interval.

If the mean number of arrivals is 10

This means that \mu = 10

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is P(X \leq 10)

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045

P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045

P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023

P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076

P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189

P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378

P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631

P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901

P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126

P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251

P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831

58.31% probability that there are no more than 10 arrivals.

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Please help !!
shutvik [7]

Given:

The limit problem is:

\lim_{x\to 3}(x^2+8x-2)

To find:

The limit of the function by using direct substitution.

Solution:

We have,

\lim_{x\to 3}(x^2+8x-2)

Applying limit, we get

\lim_{x\to 3}(x^2+8x-2)=(3)^2+8(3)-2

\lim_{x\to 3}(x^2+8x-2)=9+24-2

\lim_{x\to 3}(x^2+8x-2)=33-2

\lim_{x\to 3}(x^2+8x-2)=31

Therefore, the correct option is D.

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