All categories

3 years ago

What are the solutions to the system?

Mathematics

1 answer:

artcher [175]3 years ago

4 0

Answer:

Step-by-step explanation:

I just plugged each variable into the equation. A is the only one that works.

You might be interested in

Help please thank you

jek_recluse [69]

Answer:

the answer 1.5 if you want to know the answer see what the pattern is

4.5 - 3 = 1.5

so its going up by 1.5 :D

7 0

3 years ago

Read the following statement. If a number is divisible by 2,then it is even what is the inverse of this statement?

Lapatulllka [165]

Answer:

If it is an even number, then it is divisible by 2

Step-by-step explanation:

You flip the order of statements. One statement is "a number is divisible by 2", the other is "it is even" is the other.

7 0

3 years ago

An engineer on the ground is looking at the top of a building. The angle of elevation to the top of the building is 22°. The eng

Pavlova-9 [17]

Let the distance from the engineer to the base of the building be x, then
tan 22 = 450/x
x = 450/tan 22 = 1,113.79

Therefore the distance from the engineer to the base of the building to the nearest foot = 1,114 feet

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:

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