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Kamila [148]
2 years ago
12

If the area of a baseball diamond (shape of a square) is 8100 ft squared, how would you find the distance from 1st base to 3rd b

ase (diagonal)?
Mathematics
1 answer:
Readme [11.4K]2 years ago
5 0

Answer:

  • Exact distance = 90\sqrt{2} feet
  • Approximate distance = 127.2792 feet.

===========================================================

Explanation:

The area of the square is 8100 square feet, abbreviated ft^2.

Apply the square root to find the distance from any base to its adjacent counterpart (eg: from 1st to 2nd base). So we get sqrt(8100) = 90 ft as that side distance. Notice that 90*90 = 8100.

If you were to draw a line from 1st base to 3rd base, then you would split the square into two congruent right triangles. Each right triangle is isosceles (the two legs being 90 ft each).

Use the pythagorean theorem to find the hypotenuse.

a^2 + b^2 = c^2\\\\c = \sqrt{a^2+b^2}\\\\c = \sqrt{90^2+90^2}\\\\c = \sqrt{2*90^2}\\\\c = \sqrt{2}*\sqrt{90^2}\\\\c = \sqrt{2}*90\\\\c = 90\sqrt{2} \ \text{ ... exact distance}\\\\c \approx 127.2792 \ \text{ ... approximate distance}\\\\

The distance from 1st to 3rd base is roughly 127.2792 feet.

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Step-by-step explanation:


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Trina invited 60 people to her party and 42 said they would attend. What percent of people said they were NOT able to attend? Pl
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1. simplify 42/60 to 7/10 (42/60 divided by 2 is 21/30, 21/30 divided by 3 is 7/10)
2. if 10 x 10 = 100 than 7 x 10 = 70 so that would make it 70/100 which is 70%
so 70% is the answer

Hope this helps ;)
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Consider the probability that greater than 94 out of 153 people will not get the flu this winter. Assume the probability that a
Hatshy [7]

Answer:

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

Step-by-step explanation:

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:

p = 0.65, n = 153. So

\mu = E(X) = 153*0.65 = 99.45

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.65*0.35} = 5.9

Consider the probability that greater than 94 out of 153 people will not get the flu this winter

This probability is 1 subtracted by the pvalue of Z when X = 94. So

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

Z = \frac{94 - 99.45}{5.9}

Z = -0.92

Z = -0.92 has a pvalue of 0.1788

1 - 0.1788 = 0.8212

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

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There are no real numbers that meet your requirements.

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The two complex numbers that meet your requirement are
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