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Lostsunrise [7]
2 years ago
8

A particular state's license plates have 7 characters. Each character can be a capital letter, or a digit except for 0. How many

license plates are there in which no two adjacent characters are the same
Mathematics
1 answer:
Ede4ka [16]2 years ago
8 0

Answer:

N = 35 × 34^6

N = 54,068,154,560

Step-by-step explanation:

Given;

The license plate have 7 characters.

Each character can be a capital letter, or a digit except for 0.

There are 26 capital letters

And there are 9 digits excluding 0

The total number of possible entries in each character is;

26+9 = 35

The number of license plates in which no two adjacent characters are the same are;

For no two adjacent characters of the license plate not to be the same that is no two characters that follow each other can be the same, the number of possible entries in the adjacent characters apart from the first character would be reduced by one.

N = 35 × 34 ×34×34×34×34×34

N = 35 × 34^6

N = 54,068,154,560

Therefore, there are 54,068,154,560 possible license plates

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A small regional carrier accepted 23 reservations for a particular flight with 20 seats. 14 reservations went to regular custome
MrRissso [65]

Answer:

- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that  

P(8 arrive) = (0.56)^8 = 0.00967

- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.

Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be  

(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).  

So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079

- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,  

P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)  

= 1 - P(overbooked) - P(exactly booked)

= 1 - 0.00967 - 0.06079  

= 0.9295.

Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.

Hope that helps!

Have a great day :P

7 0
3 years ago
Which point best represents the following ordered pairs?
mrs_skeptik [129]

Answer:

B, A, E, D

Step-by-step explanation:

Hope this helps and can I have brainliest!

7 0
3 years ago
Merina is scheduled to make two loan payments to Bradford in the amount of $1,000 each, two months and nine months from now. Mer
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well, we're assuming all along that Merina owes Bradford $2000, because in the 1st scenario, she was going to pay twice $1000.

on the 2nd scenario, she'll be paying the same $2000 but split 7 months from now and then 7 months later, same 2000 bucks, at which point Bradford applied 8.5% interest.

using those assumptions, since the wording is not quite clear, we can say that Merina is simply paying 2000 bucks plus the 8.5%

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{8.5\% of 2000}}{\left( \cfrac{8.5}{100} \right)2000}\implies 170 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\stackrel{principal}{2000}~~ + ~~\stackrel{interest}{170}}{2}\implies \stackrel{\textit{two equal payments of}}{1085}

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Anuta_ua [19.1K]
I think that’s right
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A restaurant used 8.4
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