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Alika [10]
3 years ago
12

The sports field at Patton Elementary School is shaped like a rectangle.The field is 72 yards long and 46 yards wide.What is the

area of the field in square yards
Mathematics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer:

3312 square yards.

Step-by-step explanation:

Given:

The sports field at Patton Elementary School is shaped like a rectangle having:

Length = 72 yards

Breadth = 46 yards

Question asked:

What is the area of the field in square yard ?

Solution:

As here sports field at Patton Elementary School is shaped like a rectangle, thus we will use the formula ;

Area \ of\ rectangle=length \times breadth

                             =72\ yards\times46\ yards\\=3312\ square\ yards

Thus, the area of the  sports field at Patton Elementary School is 3312 square yards.

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Answer:

NORMDIST(RAND(),50,5)

Step-by-step explanation:

NORMDIST(RAND(),50,5) returns the random value from a normal distribution. The second space is meant for Mean and the third space is meant for standard deviation.

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Move the point two times.

Step-by-step explanation:

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Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 104 eligible vot
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Answer:

The probability that exactly 27 of 104 eligible voters voted is​ 0.057 = 5.7%.

Step-by-step explanation:

Binomial probability distribution

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The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed distributions 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 case, assume that 104 eligible voters aged 18-24 are randomly selected.

This means that n = 104.

Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

This means that p = 0.22

Mean and standard deviation:

\mu = 104*0.22 = 22.88

\sigma = \sqrt{104*0.22*0.78} = 4.2245

Probability that exactly 27 voted

By continuity continuity, 27 consists of values between 26.5 and 27.5, which means that this probability is the p-value of Z when X = 27.5 subtracted by the p-value of Z when X = 26.5.

X = 27.5

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

Z = \frac{27.5 - 22.88}{4.2245}

Z = 1.09

Z = 1.09 has a p-value of 0.8621

X = 26.5

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

Z = \frac{26.5 - 22.88}{4.2245}

Z = 0.86

Z = 0.86 has a p-value of 0.8051

0.8621 - 0.8051 = 0.057

The probability that exactly 27 of 104 eligible voters voted is​ 0.057 = 5.7%.

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