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3 years ago

Solve (1/6+2 2/3)x(1 1/4-1/2)

Mathematics

1 answer:

andrew11 [14]3 years ago

7 0

Answer:

Step-by-step explanation:

The expression is

(1/6+2 2/3)x(1 1/4-1/2)

Looking at the first parenthesis, (1/6 + 2 2/3), 1/6 is a proper fraction while 2 2/3 is a mixed fraction. We would convert 2 2/3 to improper fraction. It becomes 8/3. The first parenthesis becomes (1/6 + 8/3). The LCM is 6 so it becomes

(1 + 16)/6 = 17/6

Looking at the second parenthesis, (1 1/4 - 1/2), 1/2 is a proper fraction while 1 1/4 is a mixed fraction. We would convert 1 1/4 to improper fraction. It becomes 5/4. The second parenthesis becomes (5/4 - 1/2). The LCM is 4 so it becomes

(5 - 2)/4 = 3/4

The expression becomes

17/6 × 3/4 = 17/8 = 2 1/8

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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:

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