HELP PLS ITS IREADY MATH IM STUCK

Observe the sequence below 37,41,45,49

lawyer [7]

Answer:

a n = 4 n + 33

Step-by-step explanation:

n = number of jumps after the 0 term of '33'

aka: add 4 to the previous number

5 0

2 years ago

5.2.14. For the negative binomial pdf p (k; p, r) = k+r−1 (1 − p)kpr, find the maximum likelihood k estimator for p if r is know

Volgvan

Answer:

$\hat p = \frac{r}{\bar x +r}$

Step-by-step explanation:

A negative binomial random variable "is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution, this distribution is known as the Pascal distribution".

And the probability mass function is given by:

$P(X=x) = (x+r-1 C k)p^r (1-p)^{x}$

Where r represent the number successes after the k failures and p is the probability of a success on any given trial.

Solution to the problem

For this case the likehoof function is given by:

$L(\theta , x_i) = \prod_{i=1}^n f(\theta ,x_i)$

If we replace the mass function we got:

$L(p, x_i) = \prod_{i=1}^n (x_i +r-1 C k) p^r (1-p)^{x_i}$

When we take the derivate of the likehood function we got:

$l(p,x_i) = \sum_{i=1}^n [log (x_i +r-1 C k) + r log(p) + x_i log(1-p)]$

And in order to estimate the likehood estimator for p we need to take the derivate from the last expression and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\frac{x_i}{1-p}$

And we can separete the sum and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}$

Now we need to find the critical point setting equal to zero this derivate and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}=0$

$\sum_{i=1}^n \frac{r}{p} =\sum_{i=1}^n \frac{x_i}{1-p}$

For the left and right part of the expression we just have this using the properties for a sum and taking in count that p is a fixed value:

$\frac{nr}{p}= \frac{\sum_{i=1}^n x_i}{1-p}$

Now we need to solve the value of $\hat p$ from the last equation like this:

$nr(1-p) = p \sum_{i=1}^n x_i$

$nr -nrp =p \sum_{i=1}^n x_i$

$p \sum_{i=1}^n x_i +nrp = nr$

$p[\sum_{i=1}^n x_i +nr]= nr$

And if we solve for $\hat p$ we got:

$\hat p = \frac{nr}{\sum_{i=1}^n x_i +nr}$

And if we divide numerator and denominator by n we got:

$\hat p = \frac{r}{\bar x +r}$

Since $\bar x = \frac{\sum_{i=1}^n x_i}{n}$

4 0

2 years ago

At a shelter, 15% of the dogs are puppies. there are 60 dogs at the shelter. How many are puppies?

frutty [35]

So for this you need to find 15% of 60. There are a few ways to do this but I prefer to do it like this.

First, find 10% of 60. This can be found by dividing 60 by 10, because 100%/10% is 10.

This gives 10%=6

5% is half of 10%, so divide your answer for 10% (6), by 2.

This gives 5%=3

We needed 15%, so we add our 10% and 5% together, to get 9.

15% of 60 is 9, meaning there are 9 puppies.

Hope this helps!

8 0

2 years ago

Read 2 more answers

Two possible solutions of are -7 and 1. Which statement is true?

quester [9]

We can plug those values into the equation, and if the answer is incorrect, we'll know if either one is extraneous.

√11 - 2(-7) = √(-7)^2 + 4(-7) + 4

√25 = √25

5 = 5

The first solution, -7, makes the equation true, and so it is not extraneous.

√11 - 2(1) = √(1)^2 + 4(1) + 4

√9 = √9

3 = 3

The second solution, 1, makes the equation true, and so it is also not extraneous.

<h3>The correct option is D, neither solution is extraneous. </h3>

4 0

2 years ago

Briony invests some money. Interest is paid on the money at a rate of 2% per annum for the first year and 1% per annum for the s

nalin [4]

Answer:

Brionny invested £8400.

Step-by-step explanation:

Simple Interest

This is a simple interest problem.

The simple interest formula is given by:

$E = P*I*t$

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

$T = E + P$

In this question:

We divide in two intervals, the first year and the second year. We have to know that the amount after the end of the first year is the principal at the start of the second year.

First year:

Amount of 2% interest, during 1 year.

So the interest earned is:

$E = P*I*t = P*0.02*1 = 0.02P$