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scZoUnD [109]
3 years ago
11

In one semester, doug and Laura spent a combined 250 hours in the tutoring lab. If doug spent 40 more hours in the lab then Laur

a did how many hours did Laura spend in the lab
Mathematics
1 answer:
aksik [14]3 years ago
8 0

Answer:

Step-by-step explanation:

What we have is a general equation that says this in words:

Laura's hours + Doug's hours = 250 total hours

Since we don't know either person's number of hours, AND since we can only have 1 unknown in a single equation, we need to write Laura's hours in terms of Doug's, or Doug's hours in terms of Laura's.  We are told that Doug spent Laura's hours plus another 40 in the lab, so let's call Laura's hours "x".  That makes Doug's hours "x + 40".  Now we can write our general equation in terms of x:

x + x + 40 = 250 and

2x = 210 so

x = 105

Since Laura is x, she worked 105 hours in the lab and Doug worked 40 hours beyond what Laura worked.  Doug worked 145.  As long as those 2 numbers add up to 250, we did the job correctly.  105 + 145 = 250? I believe it does!!

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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}

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\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}

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