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ikadub [295]
3 years ago
11

A certain identification number is a sequence of five digits. (a) How many identification numbers are possible? Incorrect: Your

answer is incorrect. numbers (b) How many of them begin with either 023 or 003? numbers (c) How many identification numbers are possible if no two adjacent digits are the same? (For example, 123-45 is permitted, but 123-34 is not.) numbers
Mathematics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

100,000

200

65,610

Step-by-step explanation:

Given that:

Identification number = 5 digits

A.) number of identification numbers possible :

Possible digits : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Total digits possible = 10

Each digit can be any of the 10;

Hence,

10 * 10 * 10 * 10 * 10 = 100,000 possibilities

B.) How many of them begin with either 023 or 003

023 or 003

0 or 0 = 1 option

2 or 0 = 2 options

3 or 3 = 1 option

Other digits will have any of the 10 options :

1 * 2 * 1 * 10 * 10 = 200

3.) if no two digits that follow each other are the same, then only the first digit has 10 options with the other 4 each having 9

10 * 9 * 9 * 9 * 9 = 65,610

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The number of trees at the begging of the 4-year period was 2560.

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Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

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Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

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3 years ago
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