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fomenos
3 years ago
8

Which statement is true regarding the intervals where the function is increasing and decreasing? The function is increasing from

(–[infinity], 0). The function is increasing from (0, [infinity]). The function is decreasing from (–[infinity], 0). The function is decreasing from (–[infinity], [infinity]).
Mathematics
1 answer:
____ [38]3 years ago
7 0

Answer:

The function is increasing from (- infinity , 0)

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago
Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste
lidiya [134]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

For this case we have the following distribution given:

X          3      4       5        6

P(X)   0.07  0.4  0.25  0.28

We can calculate the mean with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74

In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36

And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

3 0
3 years ago
Need help with #9 please
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2 years ago
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