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KengaRu [80]
3 years ago
7

Classify the limaçon represented by the equation r=(1/2)-(1/2)sin(theta)

Mathematics
1 answer:
babunello [35]3 years ago
4 0
We know that

Limacons are in the form
r=a(+-)b*sin (theta)  or r=a(+-)b*cos (theta)

in this problem
r=a(+-)b*sin (theta)
so
is symmetric to y axis
a=1/2
b=1/2
a/b=1---------> Cardioid (hearth shaped)

see the attached figure

the answer is the option
<span>A. Cardioid </span>

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The segments shown below could form a triangle?
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Answer:

<em>Option B; False</em>

Step-by-step explanation:

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| a - b | < c < a + b,\\| c - a | < b < a + c,\\| c - b | < a < b + c, Considering - a - segment AC ( 8 units ) - b - BC ( 7 units ), -  c - BC ( 15 units )\\\\| 8 - 7 | < 15 < 8 + 7,\\1 < 15 < 15, Now 15 is not < 15, it is =, thus ;\\Solution ; False

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4 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
3 years ago
What’s 39,204 divided by 54
ozzi

Answer:

726

Step-by-step explanation:

39204/54=726

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