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loris [4]
2 years ago
13

Please Help!! DUE SOON!! I'LL MARk BRAInLiEAST!! HIGH POintS!!

Mathematics
1 answer:
Helga [31]2 years ago
8 0

Answer:

Step-by-step explanation:

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I need help on this and I need to know how to do the steps for it
cestrela7 [59]
I got you, my answer is x<4

5 0
3 years ago
Tylond puts $5,000 into a savings account that earns 4% simple interest. How much interest does he earn after
Softa [21]

Answer:

5,6241

Step-by-step explanation:

I have no idea if this is correct but I hope it is. I hope this helps.

4 0
3 years ago
Please help me I need to get this done
OverLord2011 [107]
The square root of 64 is 8 and the square root of 100 is 10.... so,

square root of 64/100 = 8/10

8/10 reduces to 4/5

Answer D.
4 0
3 years ago
Plz help due tonight!!!
jolli1 [7]

Answer:

x=75+36

x=111

Step-by-step explanation:

pls mark as brainliest

7 0
2 years ago
Read 2 more answers
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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