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Nitella [24]
3 years ago
12

Someone plz help desperateeeeee

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
3 0

We are told that Tristan is planting a garden that's square shaped and its area is 64 square meters.  Because 8 * 8 = 64, we have a square where each side is 8 meters along.

From an algebra point of view, where s is a side of the square,

s² = 64

s = 8 or s = -8 by taking square roots of both sides

s = 8 by discarding the -8 (lengths cannot be negative)

Now that we know a square's side we need its perimeter. Each side is the same, so a square's perimeter is four times a side.

4 * 8 = 32, which is our perimeter.


Therefore 32m² of material are needed.

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Answer:

a) 59.10% probability that 12 or fewer fish were caught.

b) 99.74% probability that 5 or more fish were caught.

c) 58.84% probability that between 5 and 12 fish were caught.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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

Z = \frac{X - \mu}{\sigma}

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 29, p = 0.41

So

\mu = E(X) = np = 29*0.41 = 11.89

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 2.6486

Find the following probabilities.

a) 12 or fewer fish were caught.

Using continuity correction, this is P(X \leq 12 + 0.5) = P(X \leq 12.5), which is the pvalue of Z when X = 12.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{12.5 - 11.89}{2.6486}

Z = 0.23

Z = 0.23 has a pvalue of 0.5910

59.10% probability that 12 or fewer fish were caught.

b) 5 or more fish were caught.

Using continuity correction, this is P(X \geq 5 - 0.5) = P(X \geq 4.5), which is 1 subtracted by the pvalue of Z when X = 4.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{4.5 - 11.89}{2.6486}

Z = -2.79

Z = -2.79 has a pvalue of 0.0026

1 - 0.0026 = 0.9974

99.74% probability that 5 or more fish were caught.

c) between 5 and 12 fish were caught.

Using continuity correction, this is P(5 - 0.5 \leq X \leq 12 + 0.5) = P(4.5 \leq X \leq 12.5), which is the pvalue of Z when X = 12.5 subtracted by the pvalue of Z when X = 4.5. So.

From a), when X = 12.5, Z has a pvalue of 0.5910

From b), when X = 4.5, Z has a pvalue of 0.0026.

So

0.5910 - 0.0026 = 0.5884

58.84% probability that between 5 and 12 fish were caught.

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