5. In the equation: A = 900(1.05) 10. What does 1.05 represent? **

a bag contains a-r-I-t-h-m-e-t-I-c. Amelia chooses a tile without looking and doesn’t replace it. She chooses a second tile with

olga nikolaevna [1]

<span>The first time she chooses, her chances are 2/10 that she will get an I tile. With not replacing the tile, her second chance will be 1/9. Multiply the two fractions together gives 2/90 = 1/45
(^>^)</span>

5 0

3 years ago

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Diana uses 30 grams of coffee beans to make 48 fluid ounces of coffee. When the company comes she makes 100 fluid ounces of coff

murzikaleks [220]

Sixty four fluid ounces hope that helped

7 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

<img src="https://tex.z-dn.net/?f=%20-%20y%20%3D%20%5Cfrac%7B2%7D%7B5%7Dx%20-%203" id="TexFormula1" title=" - y = \frac{2}{5}x -

raketka [301]

The first step to solving this equation is to change the signs on both sides of the equation
y = - $\frac{2}{5}$ x + 3
Since we don't know what x is but we do know why y is,, we will write our answer with the answer to y and x being left as undefined.
y = - $\frac{2}{5}$ x + 3, x ∈ R
Let me know if you have any further questions
:)

5 0

3 years ago

I WILL GIVE BRAINLIEST ANSWER!!!

Blababa [14]

Answer:

10 is unbiased

11 is biased

Step-by-step explanation:

10 is unbiased

11 is biased

Have an amazing day!

PLEASE RATE!!

7 0

2 years ago

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