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Mariana [72]
3 years ago
14

Let f be a continuous function on the closed interval [ − 2 , 5 ]. If f((-2)=-7 and f(5)=1, then the Intermediate Value Theorem

guarantees that
Mathematics
1 answer:
azamat3 years ago
6 0

Answer:

The function has at least 1 zero within the interval [-2,5].

Step-by-step explanation:

The intermediate value theorem states that, for a function continuous in a certain interval [a,b], then the function takes any value between f(a) and f(b) at some point within that interval.

This theorem has an important consequence:

If a function f(x) is continuous in an interval [a,b], and the sign of the function changes at the extreme points of the interval:

f(a)>0\\f(b) (or viceversa)

Then the function f(x) has at least one zero within the interval [a,b].

We can apply the theorem to this case. In fact, here we have a function f(x) continuous within the interval

[-2,5]

And we also know that the function changes sign at the extreme points of the interval:

f(-2)=-70

Therefore, the function has at least 1 zero within the interval [-2,5], so there is at least one point x' within this interval such that

f(x')=0

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Answer and Step-by-step explanation:

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A computer is normally $800 but is discounted to $525. What percent of the original price does Mark pay? Round to a
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Step-by-step explanation:

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The two-way table shows the number of students in a school who have rabbits and/or birds as pets:
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An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations
Hoochie [10]

Answer:

a) E(Y)= \sum_{i=1}^n Y_i P(Y_i)

And replacing we got:    

E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95

b) E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".  

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).  

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

Part a

We have the following distribution function:

Y        0         1         2       3

P(Y)  0.45    0.2    0.3   0.05

And we can calculate the expected value with the following formula:

E(Y)= \sum_{i=1}^n Y_i P(Y_i)

And replacing we got:    

E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95

Part b

For this case the new expected value would be given by:

E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)

And replacing we got

E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148

5 0
3 years ago
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