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geniusboy [140]
1 year ago
12

A license plate consists of 2 letters followed by 3 digits. How many license plates are possible if the 1st digit cannot be 0, a

nd letters and digits may repeat
Mathematics
1 answer:
zloy xaker [14]1 year ago
8 0

There are 608400 plates are possible.

If the first is A, we have 26 possibilities:

AA, AB, AC,AD,AE ...................................... AW, AX, AY, AZ.

If the first is B, we have 26 possibilities:

BA, BB, BC, BD, BE .........................................BW, BX,BY,BZ

And so on for every letter of the alphabet.

There are 26 choices for the first letter and 26 choices for the second letter. The number of different combinations of 2 letters is:

26×26=676

The same applies for the three digits.

There are 9 choices for the first as first digit cannot be 0

10 for the second,

10 for the third

9×10×10=900

So for a license plate which has 2 letters and 3 digits, there are:

676 × 900 = 608400 possibilities

To learn more about calculating possibilities from the given link

brainly.com/question/4658834

#SPJ4

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116-8= 108
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4(22464)=89856
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Answer: 140 m3

Step-by-step explanation:

Multiply the area of the trapezoid face by the length .

Since the area is 20, you multiply it by 7 to get 140

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Find each product.<br><br> (5-x2+2)(-3)<br><br> PLEASE HELP!!! ASAP!!!
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3x^2 - 21 (did you mean for the equation to be (5 - x^2 + 2) \cdot -3?)

Step-by-step explanation:

Multiplying -3 by each term:

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So the equation comes out to 3x^2 - 21 .

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A plumber has a 40-foot piece of PVC pipe. How many 11/5 foot pieces can be cut from the 40 foot piece?
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A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. His
morpeh [17]

Answer:

There is a 21.053% probability that this person made a day visit.

There is a 39.474% probability that this person made a one night visit.

There is a 39.474% probability that this person made a two night visit.

Step-by-step explanation:

We have these following percentages

20% select a day visit

50% select a one-night visit

30% select a two-night visit

40% of the day visitors make a purchase

30% of one night visitors make a purchase

50% of two night visitors make a purchase

The first step to solve this problem is finding the probability that a randomly selected visitor makes a purchase. So:

P = 0.2(0.4) + 0.5(0.3) + 0.3(0.5) = 0.38

There is a 38% probability that a randomly selected visitor makes a purchase.

Now, as for the questions, we can formulate them as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

Suppose a visitor is randomly selected and is found to have made a purchase.

How likely is it that this person made a day visit?

What is the probability that this person made a day visit, given that she made a purchase?

P(B) is the probability that the person made a day visit. So P(B) = 0.20

P(A/B) is the probability that the person who made a day visit made a purchase. So P(A/B) = 0.4

P(A) is the probability that the person made a purchase. So P(A) = 0.38

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.4*0.2}{0.38} = 0.21053

There is a 21.053% probability that this person made a day visit.

How likely is it that this person made a one-night visit?

What is the probability that this person made a one night visit, given that she made a purchase?

P(B) is the probability that the person made a one night visit. So P(B) = 0.50

P(A/B) is the probability that the person who made a one night visit made a purchase. So P(A/B) = 0.3

P(A) is the probability that the person made a purchase. So P(A) = 0.38

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.3}{0.38} = 0.39474

There is a 39.474% probability that this person made a one night visit.

How likely is it that this person made a two-night visit?

What is the probability that this person made a two night visit, given that she made a purchase?

P(B) is the probability that the person made a two night visit. So P(B) = 0.30

P(A/B) is the probability that the person who made a two night visit made a purchase. So P(A/B) = 0.5

P(A) is the probability that the person made a purchase. So P(A) = 0.38

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.5}{0.38} = 0.39474

There is a 39.474% probability that this person made a two night visit.

3 0
3 years ago
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