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3 years ago

find the domain of the function below if the domain is {-1,0,2} f(x)=x^2 -2x+3

Mathematics

1 answer:

ollegr [7]3 years ago

5 0

Answer:

The Range is {3, 6}.

Step-by-step explanation:

The correct question is

Find the range of the function below if the domain is {-1,0,2} f(x)=x^2 -2x+3

we know that

The domain represents all possible values of x.

The range represents all possible values of f(x)

Substitute all of the possible x-values (domain) into the formula to find all possible f(x) values (the range).

For x=-1

$f(-1)= (-1)^{2} - 2(-1) + 3$

$f(-1)=1+2+ 3$

$f(-1)=6$

For x=0

$f(0)= (0)^{2} - 2(0) + 3$

$f(0)=0-0+ 3$

$f(0)=3$

For x=2

$f(2)= (2)^{2} - 2(2) + 3$

$f(2)=4-4+ 3$

$f(2)=3$

therefore

The Range is {3, 6}.

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a) during the first 10 hours of growth the weight will increase 271.8% relative to the initial state ( or 2.718 g)

b) during the first 20 hours of growth the weight will increase 467.1 % relative to the initial state ( or 4.671 g)

Step-by-step explanation:

since the growing rate law is

W'(t) = 0.1 gr/hour * e^(0.1gr/hour* t) , W'(t) [gr/hour]

and following mathematical conventions: W'(t)= dW/dt

then

dW/dt=0.1 e^(0.1t)

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W = e^(0.1t) + 1 g

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W = e^(0.1 g/h*10h) + 1 g = 3.718 g/h

therefore the weight will increase

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W = e^(0.1 g/h*20h) + 1 g = 8.389 g/h

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It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.

zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

$P(X=x)=nCx*p^{x}*(1-p)^{(n-x)}$ with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

$P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)}$ with x = 0, 1

So,

$P(A)=P(X=0)+P(X=1)$

$P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}$

$P(A)=0.99^{10}+10*0.01*0.99^{9}$

$P(A)=0.9957$

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

$3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127$

Finally, we have that the probability of returning exactly one of the three packages is 1.27%

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