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3 years ago

Jika f(x-1)=2x+5 dan g1(x-2)=4x+2 maka (g•f)1(0)

1 answer:

8 0

G (x) = x + 1 (fog) (x) = x 2 + 3x + 1 ⇒ (fog) (x) = x 2 + 3x + 1 ⇒ f (g (x)) = x 2 + 3x + 1 ⇒ f (x + 1) = x 2 + 3x + 1 Eg x + 1 = p, then x = p - 1. ⇒ f (p) = (p - 1) 2 + 3 ( p - 1) + 1 ⇒ f (p) = p 2 - 2p + 1 + 3p - 3 + 1 ⇒ f (p) = p 2 + p - 1 So f (x) = x 2 + x - 1 - ->
ANSWER IS : x 2 + x - 1

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Is kilobyte bigger than megabyte.

tino4ka555 [31]

Answer:

Step-by-step explanation:

1 kilobyte = 1000 byte

1 megabyte = 1,000,000 byte

1 kilobyte = 1/1000 megabyte

Answer: no

5 0

1 year ago

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The sum of the present ages of Mark and his mother is 60 years. In 6 years she will be twice as old as he will be. Find their pr

Igoryamba

Answer:

Mark is 18; his mother is 42.

Step-by-step explanation:

In 6 years, 12 years will be added to the total of their ages, so it will be 72. Since this is the total of Mark's age and twice Mark's age, his age at that time will be 72/3 = 24. His mother will be 48 at that time.

Now, they are 6 years younger than that, so they are 18 and 42.

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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

3 years ago

Choose the correct simplification of (2x3 + x2 − 4x) − (9x3 − 3x2).

Mrrafil [7]

You start off by multiplying the second bracket by -1 to get rid of the bracket.
So,
(2x3 + x2 - 4x) - 9x3 + 3x2 =
2x3 +x2 - 4x - 9x3 + 3x2 =
(2x3 - 9x3) + (x2 + 3x2) - 4x =
-7x3 + 4x2 - 4x

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2 years ago

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solmaris [256]

Answer:first one
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2 years ago

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