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zzz [600]
3 years ago
8

An ornithologist who specializes in cockatoos in Australia decides to estimate the cockatoo population in a certain region. To d

o so, he traps 54 cockatoos and marks them. After releasing the cockatoos and waiting a bit, he traps 290 cockatoos and observes that 29 of them are marked. To the nearest whole number, what is the best estimate for the cockatoo population?
Mathematics
2 answers:
Jet001 [13]3 years ago
8 0

Answer:

540

Step-by-step explanation:

Let the estimate total population=y

Initially, out of a total of y, 54 are marked.

Then out of a sample of 290 cockatoos, 29 of them are marked.

We take the ratio of the population to the sample and do same for the number of marked in each category.

y:290 = 54: 29

\frac{y}{290}=\frac{54}{29}

Cross multiplying

y X 29 = 290 X 54

Divide both sides by 29 to obtain y.

\frac{y X 29}{29}=\frac{290 X 54}{29}

y= 54 X 10 =540

The best estimate of the cuckatoo population is 540

gogolik [260]3 years ago
4 0

Answer:

the population size is 540 cockatoos

Step-by-step explanation:

Denoting P as the total population , since the person took the 54 cockatoos and mark them , after he released them there are a proportion of marked cockatoos in the population equal to

proportion of marked cockatoos = marked cockatoos / total number of cockatoos = 54 / P

then if he takes a sample , if we assume that the marked cockatoos are well mixed around the population and each cockatoo has the same probability of being trapped  , then if we take cockatoos at random , is almost likely that he traps cockatoos in the same proportion , then

proportion of marked cockatoos = 29/290 = 54/P

P= 54 * 290 / 29 = 540 cockatoos

Note

- Mathematically , is the same that saying that each sample has the the same probability of being chosen.

- Actually if the traps 290 cockatoos out of 540 , the actual probability can be calculated through an hypergeometric distribution whose most probable value of the population size is 540 cockatoos

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7 0
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Answer:

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Step-by-step explanation:

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

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This is 1 subtracted by the pvalue of Z when X = 34. So

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8 0
3 years ago
A gardener wants to create a rectangular vegetable garden in a backyard. She wants it to have a total area of 120 square feet, a
erastova [34]
Comment
This is an area problem. The key words are 120 square feet and 12 feet longer.
And of course width is a key word when you are reading this.

Formula
Area = L * W

Givens
W = W
L = W + 12

Substitute and Solve
Area = L* W
120 = W*(W + 12)
W^2 + 12W = 120 square feet
w^2 + 12w - 120 = 0

This does not factor easily. I would have thought that a graph might help but not if the dimension has to be to the nearest 1/100 of a foot. The only thing we can do is use the quadratic formula.

a = 1
b = 12
c = - 120

w = [ -b +/- sqrt(b^2 - 4ac) ]/(2a)
w = [-12 +/- sqrt(12^2 - 4*(1)(-120)] / 2*1
w = [-12 +/- sqrt(144 - (-480)]/2
w = [-12 +/- sqrt(624)] / 2
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w = (12.979992) / 2
w = 6.489995 I'll round all this when I get done

L = w + 12
L = 6.489995 + 12
L = 18.489995

check
Area = L * W
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Answer
L = 18.489995 = 18.49 feet
W = 6.489995 = 6.49 feet

Note: in the check if you round first to the answer, LW = 120.0001 when you find the area for the check. Kind of strange how that nearest 1/100th makes a difference.
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