A word form for 10,007,000,000,000

Took a photo to use

ruslelena [56]

Answer:

x=9

<A=23

<B= 69

Step-by-step explanation:

1. 2x+5+3(2x+5)=92 2. 2x+5+6x+15=92 3. 8x+20=92 4. 8x=72 5. x=9

<A= 2(9)+5=18+5=23

<B= 3(2(9)+5)=3(23)=69

5 0

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:

$\lambda \geq 6.63835$

4 0

3 years ago

What is the highest common factor of 60 and 75

Assoli18 [71]

The factors of 60 are 60, 30, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1.The factors of 75 are 75, 25, 15, 5, 3, 1.The common factors of 60 and 75 are 15, 5, 3, 1, intersecting the two sets above.In the intersection factors of 60 ∩ factors of 75 the greatest element is 15.Therefore, the greatest common factor of 60 and 75 is 15.

8 0

3 years ago

Pittsburgh International airport had approximately 714,587 passengers in august 2009. write a number that is greater than 714,58

OLga [1]

Answer:

800,000

Step-by-step explanation:

8 0

3 years ago

What is 6/100 + 2/10

Andreyy89

Answer:

$\frac{26}{100}$

Step-by-step explanation:

$\frac{6}{100}$ can stay untouched, but $\frac{2}{10}$ has to be changed to $\frac{20}{100}$ . Now that they have they have the same denominator, you can add them. $\frac{6}{100} + \frac{20}{100} = \frac{26}{100}$ , which can be simplified, which is $\frac{13}{50}$ .

3 0

3 years ago

Read 2 more answers