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mamaluj [8]
3 years ago
12

A password is 4 characters long and must consist of 3 letters and 1 of 10 special characters. If letters can be repeated and the

special character is at the end of the password, how many possibilities are there?
Mathematics
2 answers:
Lena [83]3 years ago
8 0
175760 is the answer
Galina-37 [17]3 years ago
4 0

1st digit = 1 of 26 letters

2nd digit = 1 of 26 letters

3rd digit = 1 of 26 letters

 4th digit = 1 of 10 special characters


26 x 26 x 26 x 10 = 175,760 possibilities

 

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p = 48°

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Round the number 76.4491 to 1 decimal place
solniwko [45]

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3 years ago
getting home from trick or treat celia and emma counted their candies. half of celias candies is equal to 2/3 of emmas candies.
blagie [28]

Candies with celias and emmas is 60 and 45 respectively.

<u>Solution:</u>

Given, Getting home from trick or treat celia and emma counted their candies.  Half of celias candies is equal to 2/3 of emmas candies.  

They had a total of 105 candies altogether.  

We have to find how many candies did each of them have.

Let the number of candies with celias be n, then number of candies with emma will be 105 – n.

Now according to given condition.

\begin{array}{l}{\frac{1}{2} \times \text { celias candies count }=\frac{2}{3} \times \text { emmas candies count }} \\\\ {\rightarrow \frac{1}{2} \times n=\frac{2}{3} \times(105-n)} \\\\ {\rightarrow 3 \times n=2 \times 2 \times(105-n)} \\\\ {\rightarrow 3 \times n=2 \times 2 \times(105-n)} \\\\ {\rightarrow 3 n=4(105-n)} \\\\ {\quad \rightarrow 3 n=420-4 n} \\\\ {\rightarrow 3 n=420-4 n} \\\\ {\rightarrow 3 n+4 n=420} \\\\ {\rightarrow 7 n=7 \times 60} \\\\ {\rightarrow n=60}\end{array}

Hence, candies with celias and emmas is 60 and 45 respectively.

4 0
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Step-by-step explanation:

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8 0
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Use I = PRT to solve.
Mice21 [21]

Answer:

Time T = 15 year

Step-by-step explanation:

Given:

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270,000 = (18,000)(T)

Time T = 15 year

4 0
3 years ago
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