1. The state fair charges $6 for admission plus $2.25 for each ride. If Sarah has $30 to spend, how many rides can she go on?

Mathematics

1 answer:

umka2103 [35]2 years ago

3 0

Answer:

Step-by-step explanation:

she has 30 dollars and you take out the admission fee so you get 24 dollars. then, just divide 24 by 2.25 which will give you 10.6 since you can not go on a 0.06 of a ride then it is ten because you round to the bottom.

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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:

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