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Elina [12.6K]
3 years ago
5

How many times does the digit one appear from 101 to 200

Mathematics
2 answers:
laila [671]3 years ago
3 0
118 times
Gdhejegevehsgcsfshahs
yuradex [85]3 years ago
3 0

Answer:

Step-by-step explanation:

Let's get the answer first and then we can discuss how it is done. The answer is 119.

There is a problem. The question does not state whether or not 101 is included. If it is 119 is correct. If the question is between 101 and 200 then the answer will be 117.

How was it done? A programming language was used. So it counted by brute force.

From 101 to 120 we get

101  102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

32  Certainly there must be a quicker way

101 = 2

102 to 109 = 1 * 8 = 8

110 to 119 = 10*2 = 20  

111 has an extra 1  = 1

120 has              <u>      1</u>

Total                       32  

From here it gets a bit easier.

121                      2

122 - 129        <u>    8</u>

total                  10

130                     1

131                      2

132 - 139         <u>   8  </u>

total                  11

140 to 149        11

150 to 159        11

160 to 169        11

170 to 179         11    

180 to 189         11

190 to 199         11

total 7 * 11 + 32 + 10 = 119

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valina [46]

Answer:

23 months

Also please note that it's everytime ^t/12 not (^t)/12

Step-by-step explanation:

Principal (P) = $2700

rate = 9%

t = ?

Amount = P(1+\frac{r}{100} )^(t/12)

Amount = 2700 +500 = $3200

So substituting all our values we get :

3200 = 2700(1+\frac{9}{100})^(t/12)\\ 3200=2700(1+0.09)^(t/12)\\3200 = 2700(1.09)^(t/12)\\\frac{3200}{2700}  =1.09^(t/12)

So we use log to solve for t :

t/12=log_{1.09} (\frac{3200}{2700})\\

Solving this using a calculator we get:

1.9714966193*12

=23 months

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Answer:

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