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AveGali [126]
2 years ago
7

Team A faces loses of 5,876 points in Round 1 . How much is the loss of points for each of the 26 players of Team A if the loss

is shared equally by each player?
Mathematics
1 answer:
Svetllana [295]2 years ago
6 0

Answer:

226 points were lost per player

Step-by-step explanation:

Since we are evenly spreading the points between all 26 players, we need to solve using division.

Basically, we are dividing the 5876 points among the 26 players.

5876/26 = 226

Therefore, if the points were spread evenly among all 26 players, each player would lose 226 points.

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If a die is rolled approximately 20 times and the number of twos that come up is tallied. The probability of getting exactly four 2's are <span> 0.202. This will make the answer choice A. </span>
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2 years ago
The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th
Feliz [49]

Answer:

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

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

By the Central Limit Theorem

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

Z = \frac{30.9 - 29}{0.8341}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

X = 27.1

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

Z = \frac{27.1 - 29}{0.8341}

Z = -2.28

Z = -2.28 has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

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2 years ago
23 security systems a home security system is designed to have a 99% reliability rate. suppose that nine homes equipped with thi
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C. eight or fewer

99% means 99/100

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Find a matrix representation of the transformation L(x, y) = (3x + 4y, x − 2y).
mario62 [17]

Answer:

\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}3&1\\4&-2\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]

Step-by-step explanation:

The general matrix representation for this transformation would be:

\left[\begin{array}{cc}x&y\end{array}\right] * A = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]

As the matrix A should have the same amount of rows as columns in the firs matrix and the same amount of columns as the result matrix it should be a 2x2 matrix.

\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]

Solving the matrix product you have that the members of the result matrix are:

3x+4y = a*x + c*y

x - 2y = b*x + d*y

So the matrix A should be:

\left[\begin{array}{cc}3&1\\4&-2\end{array}\right]

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