PLEASE HELP ME ASAP

What are the x-intercept and y-intercept of the graph of y= -1/2x + 4

olga nikolaevna [1]

X intercept is 8
y intercept is 4

8 0

3 years ago

Solve for x.<br><br> 12(x−360)=120 <br><br> Enter your answer in the box.

kenny6666 [7]

12x-4320=120
12x=120+4320
12x=4440
x=370

6 0

3 years ago

Read 2 more answers

Can someone help me with these three

pav-90 [236]

#17: is 6
#18: is 30
#19: is 3.75

3 0

3 years ago

Zoe has $90 in a savings account. The interest rate is 10% per year and is not compounded. How much interest will she earn in 1

kirza4 [7]

Answer:

k,,kk,n n jn jn jmnj jj jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjuukkkjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

6 0

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:

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

4 0

3 years ago