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3 years ago

A family has two cars. The first car has a fuel efficiency of

1 answer:

8 0

The first car consumed 21 gallons while the second car consumed 49 gallons. here is the how it's done.
for the first car gallons consumed is 15/50x70= 21 gallons
the second car consumed 35/50x70=49 gallons

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Some hot dogs come in packages of 8.why would a baker of hot dog buns package 7 hot dog buns to a package

Nadusha1986 [10]

Not sure what this has to do with maths, but probably because it would mean that people would have to buy 2 packs of hot dog buns just to get 8 hot dog buns, which makes him more money?

7 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

Rename 5400 using place value chart

AlekseyPX

5,400 I think I'm right

4 0

3 years ago

If we pick two respondents at random, what's the probability they both selected purple

Mrrafil [7]

Answer:

This depends on what other colors are in the question. But put 1/x and replace the x with how many colors there are and thats thier probabilaty of choosing purple.

Step-by-step explanation:

Theres no possible way of knowing unless you already have met and or know the person so if theres green, blue, pink, yellow, and purple, theres 5 colors and purple is 1 out of them therefore, 1/5

7 0

3 years ago

Please help!!! I will make you brainliest I need this and I also don't understand

Naily [24]

3 it makes 2 parallelograms so it is 3 but if you need more help Comment on this answer

3 0

2 years ago