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lubasha [3.4K]
2 years ago
5

Given the points (0,−3) and (5,-6) and on a line, find its equation in the form y=mx+b

Mathematics
1 answer:
Andrew [12]2 years ago
4 0

Answer:

Step-by-step explanation:

y=mx+b

b=-3 (y intercept)

Use equation y2-y1/x2-x1

-3/5= m

y=-3/5x-3

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If time is four hours from now, the time left till midnight would be quarter that if it is one hour from now. What time is it no
lapo4ka [179]

Answer:

The time now is 7 pm

Step-by-step explanation:

Suppose that now is the time T.

We know that:

"if time is four hours from now, the time left till midnight would be a quarter that if it is one hour from now".

Then:

if we define midnight as 12, and we assume that T is in the pm range.

Then the "time left till midnigth, assuming that the time is four hours from now" will be written as (12 - (T + 4))

With this in mind, we can write the problem as:

12 - (T + 4) = (1/4)*( 12 - (T + 1))

Now we can solve this for T.

12 - T - 4 = (1/4)*(12 - T - 1)

8 - T = (1/4)*(11 - T)

4*(8 - T) = 11 - T

32 - 4*T = 11 - T

32 - 11  = -T + 4*T

21 = 3*T

21/3 = T

7 = T

Then T = 7 pm

The time now is 7 pm

7 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}

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Answer:

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Answer:

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