How do you do -4=-8(x-6)+4(x-8)

Suppose that the populations of the United States and China both increase by 12 million people in one year. What would be the re

Jlenok [28]

Question:

Suppose that the populations of the United States and China both increase by 12 million people in one year. What would be the relative change for each country? Answer to the nearest tenth of a percent (1 decimal place). Make sure that you fill in each answer area before checking "How Did I Do?".

If you assume that the U.S. population is about 300 million, an increase of 12 million would result in a relative change of Number %.

If you assume that China has about 1 billion people, an increase of 12 million would result in a relative change of Number %.

Answer:

The relative change of the United States and China is 4% and 1.2%, respectively.

Step-by-step explanation:

Given

Represent population with p. So, we have:

United States

$\triangle p = 12\ million$

$p_1 =300\ million$

China

$\triangle p = 12\ million$

$p_1 =1\ billion$

Required

Determine the relative change for each country

Relative change, C is calculated as:

$C = \frac{\triangle p}{p} * 100\%$

For United States

$C = \frac{12\ million}{300\ million} * 100\%$

$C = \frac{12}{300} * 100\%$

$C = \frac{12 * 100}{300}\%$

$C = \frac{1200}{300}\%$

$C = 4\%$

For China

$C = \frac{12\ million}{1\ billion} * 100\%$

$C = \frac{12}{1000} * 100\%$

$C = \frac{12* 100}{1000} \%$

$C = \frac{1200}{1000} \%$

$C = 1.2\%$

The relative change of the United States and China is 4% and 1.2%, respectively.

3 0

2 years ago

If five friends split a $98.25 bill, about how much should each friend pay?

zavuch27 [327]

Answer:

about $20 because $98.25/5=$19.65

Step-by-step explanation:

7 0

2 years ago

PLEASE HELP!! DUE SOON

zhenek [66]

Answer:

C is false as the other three are 100% correct statements. I dont know what is linear growth and exponential growth but I know that the other 3 statements are correct so the only 1 left C must be false.

4 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:

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