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aleksley [76]
3 years ago
5

What is the area of the figure below?

Mathematics
1 answer:
Marat540 [252]3 years ago
4 0

Answer:

A

Step-by-step explanation:

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Aden spent $10 on a mug and then sold it, making 20% profit, how much did he sell the mug for
NikAS [45]

Answer: $2

Step-by-step explanation: 20 percent of 10 is 2

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Juan wrapped 2 presents every 16 hours. At that rate, how long, in
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Answer:

Since it took Juan 16 hours to wrap 2 presents, it will take Juan 32 hours.

1 present- 8 hours

2 present- 16 hours

3 present- 24 hours

4 present- 32 hours

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2 years ago
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The library is 1.75 miles directly north of the school. The park is 0.6 miles directly south of the school. How far is the libra
kumpel [21]
If something is placed the way these places are(directly NORTH and the park is SOUTH) all you have to do is add your two values, lining up your decimal points. 1.75
            +.6
            2.35
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what is the vertical change shown on the graph? give the answer as a decimal rounded to the nearest tenth, if necessary
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Answer:

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8 0
3 years ago
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
3 years ago
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