I need to know this answer I've worked it out but not sure

Mathematics

2 answers:

Ksivusya [100]3 years ago

8 0

= r^-5 s^4t ^4
= s^4 t^4 / r^5

answer

answer is B, s^4 t^4 / r^5

Angelina_Jolie [31]3 years ago

7 0

$\cfrac{r^{-3}s^5t^2}{r^2st^{-2}} = \cfrac{s^4*t^2*t^2}{r^2*r^3} =\cfrac{s^4t^4}{r^5}$

Answer is B.

You might be interested in

1Y, the number of accidents per year at a given intersection, is assumed to have a Poisson distribution. Over the past few years

miss Akunina [59]

Answer:

The probability that the intersection will come under the emergency program is 0.1587.

Step-by-step explanation:

Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.

Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.

For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.

Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).

$W = \frac{X-\mu}{\sigma} = \frac{X-3}{1/2} = 2x-6$

The values of $\phi$ , the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program