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IgorLugansk [536]
3 years ago
13

(5,9)(-2,9) please help find the slope

Mathematics
2 answers:
Delicious77 [7]3 years ago
8 0
Slope formula: 
\frac{y_2-y_1}{x_2-x_1} =slope


We are given 2 coordinates: 
<span>(5,9) and (-2,9)
</span>x1,y1       x2,y2

\frac{9-9}{-2-5} = \frac{0}{-7}

0/-7=0 
Slope=0 


statuscvo [17]3 years ago
6 0
9-9/-2-5
0/-7
slope=0/-7
= 0
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Fill in the blank. Given O below, you can conclude that OD is congruent to ___________.
Debora [2.8K]

The circle with center O has two chords AC and EF which are of same length 9.07.

OD and OB are the two perpendiculars drawn from the center O to the two chords AC and EF .It represents the distance of the chords from the centre.

The circle theorem states: congruent chords are equidistant from the center.

OD is congruent to OB.

Option A is the right answer.

4 0
3 years ago
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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
3 years ago
Work out and simplify where possible<br><br>7/15+2/15​
alisha [4.7K]

Answer:

3/5

Step-by-step explanation:

7/15 + 2/15 = 9/15

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6 0
2 years ago
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mylen [45]

Answer:

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Step-by-step explanation:

Suppose the formula is ...

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__

Here's a way to solve these equations.

Subtract the first equation from the second:

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Subtract the fourth equation from the fifth:

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Then substituting into the 4th equation to find b, we have ...

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and ...

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  5 = c

The formula is ...

  f(n) = -n^2 -3n +5

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If the area of a rectangle 24a2b and the length is 8ab2, what would the width of the rectangle, given that width is found by div
slava [35]
Area of a rectangle: A = 24 a² b
Length: L = 8 a b²
Width: W = A / L = 24 a² b / 8 a b²
Answer:  W = 3 a / b 
5 0
3 years ago
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