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elena-14-01-66 [18.8K]
3 years ago
6

Try it

Mathematics
1 answer:
Valentin [98]3 years ago
4 0

Answer:

(x,y) + (x-4, y)

Step-by-step explanation:

horizontally means it would be on the x axis traveling to the left would be your subtracting there fore the only answer which comes with a x-4 is (x,y) + (x-4, y)

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N(-2,-2) After a rotation 90 degrees counterclockwise
RUDIKE [14]

Answer: (2,-2)

Step-by-step explanation:

Rotating 90 degrees counterclockwise (about the origin) maps (x,y) onto (-y,x), so the answer is <u>N'(2,-2)</u>

6 0
2 years ago
How do I solve 28.8=18x
kolezko [41]
28.8 = 18x....divide both sides by 18, eliminating the 18 on the right side
28.8 / 18 = x
1.6 = x <==
3 0
2 years ago
What is the area of the figure shown below? giving brainly!!
siniylev [52]

Answer:

4073.76

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
What is the value of y when x=3 in this equation y=2x+2
Ksenya-84 [330]

Answer:

8

Step-by-step explanation:

if you substitute 3 for x the equation becomes y=2(3) + 2

using order of operations you multiply 3 by 2 and then add 2 which will give you 8

4 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
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