3 years ago

HELP ASAP!!!!

Mathematics

1 answer:

nydimaria [60]3 years ago

3 0

The following is true:

H(x) has a constant output of -2.50

G(x) is greater than -2.50 for x values less than -1

The input value for which g(x)=h(x) is between -1 and 0

1, 3, and 5 is true

A, C, E is true

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$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

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For this case we have the following distribution given:

X 0 1 2 3 4 5 6

P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02

For this case we need to find first the expected value given by:

$E(X) = \sum_{i=1}^n X_i P(X_I)$

And replacing we got:

$E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61$

Now we can find the second moment given by:

$E(X^2) =\sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97$

And the variance would be given by:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

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