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timofeeve [1]
3 years ago
6

What is the first step to finding the equation of a line when 2 points are known?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
5 0
The first step to finding the equation of a line when 2 points are known....
the first step would be to use the slope formula to find the slope of the line
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Which of the following represents the prime factortzation of 72?
rjkz [21]

Answer:

2³ x 3²

Step-by-step explanation:

The prime factorization of a number is the process of breaking down a number to a set of prime numbers that their product will give the original number:

          72

      2   x    36

      2 x 2 x 18

      2 x 2 x 2 x 9

       2 x 2  x 2 x 3 x 3

So;

  The prime factorization is 2³ x 3²

8 0
3 years ago
F(x) = 722<br> g(x) = 2.c + 1<br> Find ( 4) (2<br> (2). Include any restrictions on the domain.
balandron [24]

Answer:

sorry this question is tough

6 0
2 years ago
Un<br> How many solutions does this system have?<br> 12x-4y=-8<br> y=3x+2
Talja [164]

Answer:

Infinite

Step-by-step explanation:

Get both in the form y = first.

12x - 4y = -8

-4y = -8 - 12x

y = 2 + 3x

y = 3x + 2

These are the same line so all points are the same so that means there are infinite answers.

If ONLY the number in front of the x was the same there would be 0 answers, they would be parallel lines, and if the number in front of the xs were different there would only be one answer.  

3 0
2 years ago
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)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
2 years ago
Solve the following system of equations using the elimination method. 6x – 5y = –1 –6x – 4y = 10
Elis [28]

Answer:

x= -1 y= -1

Step-by-step explanation:

5 0
3 years ago
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