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Romashka-Z-Leto [24]
9 months ago
8

in a certain country license plates consist of zero or one digit followed by four or five uppercase letters from the roman alpha

bet. how many different license plates can the country produce?
Mathematics
1 answer:
algol139 months ago
3 0

118,813,760 different combination license plates can the country produce

<h3 /><h3>permutations and combinations:</h3>

permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. By considering the ratio of the number of desired subsets to the number of all possible subsets for many games of chance in the 17th century, the French mathematicians Blaise Pascal and Pierre de Fermat gave impetus to the development of combinatorics and probability theory.

given that

in a certain country license plates consist of zero or one digit followed by four or five uppercase letters from the roman alphabet

X = 10 × 26 × 26 × 26 × 26 × 26

  = 10 × 11881376

  = 118,813,760

118,813,760 different combinations license plates can the country produce

To learn more about combinations:

brainly.com/question/19692242

#SPJ4

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What is 2^3 x 2^7= equivalent to
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Answer:

3.940201X 10^115

Step-by-step explanation:

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7 0
2 years ago
In a circle with a radius of 27 2/5 in., an arc is intercepted by a central angle of 7π/4 radians.
VikaD [51]
Arc Length = radius * Angle in radians
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Source:
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5 0
3 years ago
Read 2 more answers
The weight of people in a small town in Missouri is known to be normally distributed with a mean of 186 pounds and a standard de
OleMash [197]

Answer:

the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

Step-by-step explanation:

The summary of the given statistical data set are:

Sample Mean = 186

Standard deviation = 29

Maximum capacity 3,417 pounds or 17 persons.

sample size = 17

population mean =3417

The objective is to determine the probability  that a random sample of 17 persons will exceed the weight limit of 3,417 pounds

In order to do that;

Let assume X to be the random variable that follows the normal distribution;

where;

Mean \mu = 186 × 17 = 3162

Standard deviation = 29* \sqrt{17}

Standard deviation = 119.57

P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})

P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})

P(X>3417) = P(Z>\dfrac{255}{119.57})

P(X>3417) = P(Z>2.133)

P(X>3417) =1- 0.9834

P(X>3417) =0.0166

Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

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3 years ago
Somebody can help me please
sergey [27]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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