Which equation is the equation of the line, in point-slope form, that has a slope of 2 and passes through the point (−8, 1) ?

Goryan [66]

78 x 100 = 7800

How many times does 190 go into 780? 
≈ 4....190 x 4 = 760 

780 - 760 = 20...bring down the extra 0 to make it 200. 

How many times does 190 go into 200? 
≈ 1...subtract 190 from 200 to get a remainder of 10 

190 ÷ 7800 ≈ 41

Hope I Helped You!!! :-)

Have A Good Day!!!

5 0

3 years ago

a hockey season ticket holder pays 72.48 for her tickets plus 6.00 for a program each game each game a second person pays 18.08

tiny-mole [99]

If I understand the question correct, the hockey season ticket holder paid $72.48 for all of the tickets at once, and then $6.00 for each game thereafter. The second person pays $18.08 for a ticket per game.

So then let g = the number of games. The problem can be written as a system of equations:

Person 1 = 72.48 + 6g
Person 2 = 18.08g

We are looking for when the number of games for both are the same. To do this, just set them equal to each other and solve for g.

72.48 + 6g = 18.08g
Subtract 6g:
72.48 = 12.08g
Divide by 12.08:
g = 6

So after 6 games, the two would have paid the same price.

3 0

4 years ago

The economics department has 40 applicants for their phd program and only 10 openings. how many different combinations of applic

lana66690 [7]

In a sequence of events, the total possible number of ways all events can performed is the product of the possible number of ways each individual event can be performed. Since there are 10 openings

P = 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 = 3.07599 x 10^15 different combinations

5 0

3 years ago

Find the least common multiple of 12 and 48. 144 48 32 96 Help

Brilliant_brown [7]

Answer:

Step-by-step explanation:

48

4 0

3 years ago

Read 2 more answers

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