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3 years ago

What is the value of [-4.6]

Mathematics

2 answers:

solniwko [45]3 years ago

7 0

<h2>Answer:</h2>

The value is:

-4

<h2>Step-by-step explanation:</h2>

We are asked to find the value of the ceiling function: $\left \lceil -4.6 \right \rceil$

As we know that the ceiling function always occupy the higher value in integers.

i.e. the ceiling function act as follows:

it takes value 0 when -1< x≤0

1 when 0 < x ≤ 1

2 when 1 < x ≤2

and so on.

As we know that:

-4.6 lie between -5 and -4.

Hence, we have:

$\left \lceil -4.6 \right \rceil=-4$

NeTakaya3 years ago

5 0

Answer: -4

Step-by-step explanation:

The ceiling function (also known as the least integer function) is written as

$f(x) = [x]$

It gives the smallest integer greater than or equal to x .

For example : [5.6]=6

or [-1.9]= -1 [∵- 1 > -1.9 ]

To find : The value of [-4.6]

Clearly , [-4.6] is written in ceiling function notation.

Then, [-4.6] = smallest integer greater than or equal to -4.6

= -4 [∵ -4>-4.6]

Hence, the value of [-4.6] = -4

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Simplify the following problem.<br> (-8+61) +(6+61)

Travka [436]

Answer: -8+61 = 53 and 6+61 = 67

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3 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:

$\lambda \geq 6.63835$

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Read 2 more answers

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