Find the value of x to the nearest tenth (look at picture)

Anastasy [175]

The answer I came up with was ×^6-4. Just rewrite it. I hope this was helpful.

3 0

3 years ago

Determine the number of outcomes in the event. Decide whether the event is a simple event or not. A computer is used to select r

Oliga [24]

The probability of selecting a number greater than 8 will be $p=\dfrac{1}{9}$ .

<h3>What is probability?</h3>

Probability is defined as the ratio of the number of favourable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.

The number of outcomes = { 1,2,3,4,5,6,7,8,9 } = Counts = 9

Favourable outcomes = { 9 } = counts = 1

The probability will be given as:-

$P = \dfrac{1}{9}$

Therefore the probability of selecting a number greater than 8 will be $p=\dfrac{1}{9}$ .

To know more about probability follow

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7 0

2 years ago

If ∠2 and ∠3 are complementary angles and m∠2 = 24°, find m∠3.

Yuliya22 [10]

Answer:

sorry dont know

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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:

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