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Nastasia [14]
3 years ago
12

2000 soldiers stand in a row. Beginning from the left, each soldier calls out a number, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, and

so on. Beginning from the right, each soldier calls out a number 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, and so on. How many of these soldiers call out a number 3 when beginning from the left and right??
Mathematics
1 answer:
Cloud [144]3 years ago
8 0

Answer:

The number of soldiers that call out a number 3 is 666 soldiers

Step-by-step explanation:

The parameters given are;

Number of soldiers = 2000

Soldiers calling from the left = 1, 2, 3,.....

Soldiers calling from the right = 1, 2, 3,.....

Therefore, since the number soldiers calling out the number 3 are 1 in 3 from the left and 1 in 3 from the right, we split the soldiers into 2 groups of 1000, with one group calling from left and the other group calling from the right;

The number of 3s in called out in the first group from the left is therefore;

1000/3 = 333.33 which is 333 soldiers  or from the 3rd soldier to the 999th soldier

Similarly the number of 3s called out from the second group = 333

Hence the total number of soldiers that call out a number 3 = 333 + 333 = 666 soldiers.

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Answer:

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And we can find this probability with this difference and using the normal standard table:

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Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

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We are interested on this probability

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And the best way to solve this problem is using the normal standard distribution and the z score given by:

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If we apply this formula to our probability we got this:

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8 0
3 years ago
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bazaltina [42]
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4 0
3 years ago
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Theorem: A line parallel to one side of a triangle divides the other two proportionately.
Kisachek [45]

Answer:

Segment BF = 16

Step-by-step explanation:

The given theorem states that a line parallel to one side of a triangle divides the other two sides proportionately

The given theorem is the Triangle Proportionality Theorem

According to the theorem, given that segment DE is parallel to segment BC, we have;

\dfrac{AD}{BD} = \dfrac{AE}{EC}

Therefore;

BD = \dfrac{AD}{\left(\dfrac{AE}{EC} \right) }  = AD \times \dfrac{EC}{AE}

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Similarly, given that EF is parallel to AB, we get;

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