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aliya0001 [1]
2 years ago
9

The ratio x : 8 is equivalent to the ratio 8 : x^2. What is the value of x?

Mathematics
2 answers:
dimulka [17.4K]2 years ago
7 0

Answer:

x=1

Step-by-step explanation:

VashaNatasha [74]2 years ago
3 0
X=4

4:8 is equivalent to 8:16, both being simplified to 1:2
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(07.05 HC)
lianna [129]

Answer:

c

Step-by-step explanation:

2×8=16

not 10 that is why c is correct

8 0
3 years ago
to communication companies offer calling plans with company X it cost $0.30 to connect and then $0.03 for each minute with compa
mario62 [17]

Answer:

$0.30 + $0.03n > $0.02 + $0.02n

Step-by-step explanation:

company x = $0.30 + $0.03n

company y = $0.02 + $0.02n

$0.30 + $0.03n > $0.02 + $0.02n

4 0
2 years ago
A theater seats 378 people. There are two shows each weekend for four weekends. If all of the seats are sold for all performance
Lapatulllka [165]
This may not be right but 378 x 4 = 1,512
3 0
3 years ago
Read 2 more answers
PLEAS HELP <br><br>The measures 2^3 , 4, and 2^7 create a right triangle.<br>True<br>false
natali 33 [55]

Answer: true

Step-by-step explanation: they are all equivalent

6 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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