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hichkok12 [17]
2 years ago
9

Students in a marching band want to line up for their performance. The problem is that when they Line up in twos there is 1 left

over. When they line up in threes there are 2 left over. When they line up in fours there are 3 left over. When they line up in fives there are 4 left over. When they line up in sixes there are 5 left over. When they line up in sevens there are no students left over. How many students are there?
Mathematics
1 answer:
olganol [36]2 years ago
5 0

Answer:

x = 119

Step-by-step explanation:

Solution:-

- The number of students in a marching band = x

- When they Line up in "twos" there is 1 left over, That if we mathematically  express it:

                   Division: x / 2 , Remainder = 1

- When they Line up in "threes" there is 2 left over, That if we mathematically  express it:

                   Division: x / 3 , Remainder = 2

- When they Line up in "fours" there is 3 left over, That if we mathematically  express it:

                   Division: x / 4 , Remainder = 3

When they Line up in "fives" there is 4 left over, That if we mathematically  express it:

                   Division: x / 5 , Remainder = 4

When they Line up in "sixes" there is 5 left over, That if we mathematically  express it:

                   Division: x / 6 , Remainder = 5

When they Line up in "sevens" there are no left over, That if we mathematically  express it:

                   Division: x / 7 , Remainder = 0

- It means that the total number of "x" students are perfectly divisible by 7. If it is not divisible by 2, then it is an odd number.

- So,

                  x = 7 * a

Where,       x > 7 and exclude derivative multiples of (5, 4 , 6)

- So from trial and error, a = 17

                  x = 119  

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Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300.
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Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

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.

In this problem, we have that:

\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10.

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when X = 500. So

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

Z = \frac{500 - 465}{10}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.

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