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nalin [4]
1 year ago
12

Automobile license plates for a state consist of four letters followed by a dash and two single digits. How many different licen

se plate combinations are possible if exactly one letter is repeated exactly once, but digits cannot be repeated
Mathematics
1 answer:
Sauron [17]1 year ago
4 0

The number of different license plate combinations that are possible if exactly one letter is repeated exactly once, but digits cannot be repeated is 8,424,000.

<h3>What is combination?</h3>

A combination is just a mathematical technique for determining the number of potential arrangements in a set of objects where the order of a selection is irrelevant.

You can choose the components in any order in combinations. Permutations and combinations are often mistaken.

Now according to the question,

Possible letter combinations

Choose any letter and make it a repeat letter = 26 ways

But, there are ⁴C₂ = 6 spots available for the identical letters.

And there are (25)×(24) other methods for selecting the other two letters.

The total amount of "words" equals ⁴C₂ × 26 × 25 × 24 = 93600.

Furthermore, because the numerals cannot be repeated = 10 × 9 = 90

So, the total number of choices = 93600 × 90 = 8,424,000

Therefore, the total combinations in which the letters can be chosen for the license plates is  8,424,000.

To know more about the combination, here

brainly.com/question/11732255

#SPJ4

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Answer:

3.29568453×10^17

Step-by-step explanation:

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Is (8+√2)(8-√2) Rational or irrational no need to explain why
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An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and
egoroff_w [7]

Answer:

a) P(140

P(0.036

P(0.036

b) P(140< \bar X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

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:

X \sim N(139,28.1)  

Where \mu=139 and \sigma=28.1

We are interested on this probability

P(140

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

z=\frac{x-\mu}{\sigma}

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

P(140

And we can find this probability like this:

P(0.036

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(0.036

Part b

For this case we select a sample size of n =32. Since the distribution for X is normal then the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=139, \frac{\sigma}{\sqrt{n}}=\frac{28.1}{\sqrt{32}}=4.97)

And the new z score would be:

Z=\frac{\bar X -\mu}{\sigma_{\bar x}}

P(140< \bar X

P(140< \bar X

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