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3 years ago

10

Keith buys 45 tickets for rides at the amusement park. Each ride requires 6 tickets to ride. If Keith already has gone on x ride

s, which expression is equivalent to
the number of tickets Keith has left?
A:45-6x
B:6(x-45)
C:6(45-x)
D:x(45+6)

1 answer:

Lynna [10]3 years ago

3 0

A is the correct answer if I did it right

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for every $50 of candy you sell, you get to keep 7% of it. how much commission do you make if you sell $200 worth of candy?

____ [38]

You'll make 14 dollars.

7% of 50 is 3.5.

If you add 3.5 four times you'll get 14.

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3 years ago

Read 2 more answers

May you run 7 miles in 50 minutes at the same rate how many miles would she run in 75 minutes

Tcecarenko [31]

ok well you would set up an equation 7miles/50 mins=xmiles/75mins then you would cross multiply 7*75=50x

then you simplify 525=50x

then divide both sides by 50 and you end up with 10.5

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3 years ago

A computer system uses passwords that are exactly six characters and each character is one of the 26 letters (a–z) or 10 integer

ELEN [110]

Answer:

The number of hits would follow a binomial distribution with $n =10,\!000$ and $p \approx 4.59 \times 10^{-6}$ .

The probability of finding $0$ hits is approximately $0.955$ (or equivalently, approximately $95.5\%$ .)

The mean of the number of hits is approximately $0.0459$ . The variance of the number of hits is approximately $0.0459\!$ (not the same number as the mean.)

Step-by-step explanation:

There are $(26 + 10)^{6} \approx 2.18 \times 10^{9}$ possible passwords in this set. (Approximately two billion possible passwords.)

Each one of the $10^{9}$ randomly-selected passwords would have an approximately $\displaystyle \frac{10,\!000}{2.18 \times 10^{9}}$ chance of matching one of the users' password.

Denote that probability as $p$ :

$p := \displaystyle \frac{10,\!000}{2.18 \times 10^{9}} \approx 4.59 \times 10^{-6}$ .

For any one of the $10^{9}$ randomly-selected passwords, let $1$ denote a hit and $0$ denote no hits. Using that notation, whether a selected password hits would follow a bernoulli distribution with $p \approx 4.59 \times 10^{-6}$ as the likelihood of success.

Sum these $0$ 's and $1$ 's over the set of the $10^{9}$ randomly-selected passwords, and the result would represent the total number of hits.

Assume that these $10^{9}$ randomly-selected passwords are sampled independently with repetition. Whether each selected password hits would be independent from one another.

Hence, the total number of hits would follow a binomial distribution with $n = 10^{9}$ trials (a billion trials) and $p \approx 4.59 \times 10^{-6}$ as the chance of success on any given trial.

The probability of getting no hit would be:

$(1 - p)^{n} \approx 7 \times 10^{-1996} \approx 0$ .

(Since $(1 - p)$ is between $0$ and $1$ , the value of $(1 - p)^{n}$ would approach $0\!$ as the value of $n$ approaches infinity.)

The mean of this binomial distribution would be: $n\cdot p \approx (10^{9}) \times (4.59 \times 10^{-6}) \approx 0.0459$ .

The variance of this binomial distribution would be:

$\begin{aligned}& n \cdot p \cdot (1 - p)\\ & \approx(10^{9}) \times (4.59 \times 10^{-6}) \times (1- 4.59 \times 10^{-6})\\ &\approx 4.59 \times 10^{-6}\end{aligned}$ .

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