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3 years ago

118.380 round to the nearest hundred

1 answer:

swat323 years ago

4 0

The hundreds column is the 3rd digit to the left of the decimal point, in this case it's the 1.

Look at the number 1 to the right of it. If its 5 or above, you round the number up, if it's below 5 you leave it the same.

In this case the number after the 1 is another 1, so you don't round it up.

118.380 to the nearest hundred is 100

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Write a line in slope-intercept form that has a m=4 and b=-2.

Arlecino [84]

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2 years ago

A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the

olchik [2.2K]

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

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Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

$F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}$

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.

$F(X) = P ( 3=$

- The Expected Value can be determined by sum to infinity of CDF:

E(X) = Σ ( 1 - F(X) )

$E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}$

E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]

E(X) = 1

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Answer:

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