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3 years ago

F(x)= x^2 - 2x + 7 Find F(-1)

Mathematics

2 answers:

Leni [432]3 years ago

8 0

Answer:f(1)=-4, f(1)=1

Just substitute the value to the x, Like this.

F(x)=x²+2x-7

F(1)=(1)²+2(1)-7

=1+2-7

=3-7

=-4

Step-by-step explanation:

Levart [38]3 years ago

6 0

Answer:

f(- 1) = 10

Step-by-step explanation:

Substitute x = - 1 into f(x) , that is

f(- 1) = (- 1)² - 2(- 1) + 7 = 1 + 2 + 7 = 10

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Match the formula to the correct figure.

ss7ja [257]

It is asking you to move the right formula with its match.

Formula for a cylinder: π*R^2*H
So the first option goes to the first shape.

Example:
Radius=5
Length=10
π≈3.14
3.14*5²*10
=725 cm³ for the answer.

Formula for a square: l*w*h
So the 4th option goes with the 2nd shape.

Example: l=5 w=5 h=5
(Length * Width * Hieght)

5*5*5=125 for the answer.

Formula for a cone is: 1/3π*r^2*h
So the second option goes to the 3rd shape.

Example: r=3 h=11
π≈3.14
$\frac{1}{3} *3 ^{2} *11=33$
33 is your answer.

Formula for the pyramid would be: 1/3*b*h
So the 3rd option would go to the 4th shape.

Example:
b=2 h=10
$\frac{1}{3} *2*10=6.666667$
Rounded: 6.6

Hope this helps!~

(Look at the attachment)

6 0

3 years ago

Read 2 more answers

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$