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Andrej [43]
3 years ago
6

Q8 Q2.) What are the​ lot's dimensions?

Mathematics
1 answer:
Triss [41]3 years ago
7 0
Equation for perimeter is 2L +2W

perimeter of the lot is 390 ft

 so we have 390 = 2L +2w

divide all terms by 2:

195 = L + W

rewrite this as L = 195-W  ( eq.1)

 cost of fence for the length = 36 per foot so we have 36L
cost of fence for width = 6 per foot so we have 6W

the fence was 1 length and 2 sides
 so we have 36l + 6w +6w = total cost

total cost is 4980

so we have 4980 = 36L + 6w +6w

combine like terms:

4980 = 36L +12w

replace L with eq1 from above:

4980 = 36(195-w) +12w

distribute:

4980 = 7020 -36w +12w

combine like terms:

4980 = 7020-24w

subtract 7020 from both sides:

-2040 = -24w
 divide both sides by -24:

-2040 / -24 = 85
width = 85 feet

using eq1 L = 195 - 85

Length = 110 feet

lond = 110 feet
 short = 85 feet








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Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

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:

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The standard deviation of the binomial distribution is:

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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:

\mu = E(X) = np = 100*0.23 = 23

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.23*0.77} = 4.21

What is the probability that under 30% of the dogs are emotional support trained?

30% of 100 is 0.3*100 = 30

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

Z = \frac{30 - 23}{4.1}

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So there is a 95.64% probability that under 30% of the dogs are emotional support trained

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Answer:

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