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Xelga [282]
3 years ago
9

A geometric figure is a set of a finite number of points

Mathematics
1 answer:
horsena [70]3 years ago
8 0
Can you buttress your point?

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The population of Virginia is 30% less than the population of Georgia. If
Advocard [28]

7,150,000

The population of Georgia is 10,500,000.

The population of Virginia is 30% less than that.

So the population of Virginia is 10,500,000 - 30% which equals 7,150,000

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3 years ago
A sphere has a radius of 9 in, which means its volume is 972π in3. The sphere is dilated by 2⁄3. What is the volume of the new s
swat32

Answer:

08-Oct-2009 — If a spherical balloon has a volume of 972 pi cubic centimeters, what is ... is 4.pi.r^2 (its area remember not volume) as 4/3.pi.r^3=972pi r=9 ... and it is good to know the squares of the next 10 numbers.

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3 years ago
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What percentage of the survey respondents did not like either soccer or volleyball?
Snezhnost [94]
Beats me man not enough information
3 0
3 years ago
Math, i can give you brainliset just help me with the two questions :)
Kay [80]

Answer:

2. the interquartile range is 20

3. 80 is the median.

Step-by-step explanation:

have a nice day.

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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