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3 years ago

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A local college student will be randomly selected to attend a leadership conference in Washington, D.C. There are 2110 local col

lege students, of which 40% are male. 610 of the college students are graduate students, and half of the graduate students are female. Which of the following best represents the probability that the selected college student will be a male student or a graduate student

1 answer:

Lemur [1.5K]3 years ago

6 0

The Apex answer is 54%. Hope this helps! ^ your answer would have been right but you forgot to multiply 25% with the second half of the asked question.

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enyata [817]

Answer:

64 players are left-handed

Step-by-step explanation:

80/100 x 80

5 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

WILL GIVE BRAINLIEST!!!! LOOK BELOW TO ANSWER MY QUESTION.

galben [10]

1.20(35)

this would be the same as :
(1 + 0.2)* 35 <==
or it also could have been 35 + 0.20(35)

7 0

4 years ago

Someone plz help me!!!

Over [174]

$w(x)=|-x-3|\\\\q(x)=|x|+3\\\\h(x)=|x-3|\\\\f(x)=|x+3|-3$

8 0

4 years ago

Mrs. Snell's math test has twenty questions and is worth 100 points. The test consists of multiple choice questions worth 3 poin

Irina18 [472]

Answer:

The number of multiple-choice questions are 15.

Step-by-step explanation:

Mrs. Snell's math test has twenty questions and is worth 100 points.

The test consists of multiple-choice questions and short answer questions.

Let M represents the Multiple choice questions.

Let S represents the short answer question.

It is given that the total number of question are 20.

$S+M=20$ ...(1)

Also, multiple-choice questions worth 3 points each, and short answer questions worth 11 points each. The total worth of the test is 100 points.

$11S+3M=100$ ...(2)

Multiply equation (1) by 3 and subtract from equation (2).

$11S+3M=100\\3S+3M=60\\\underline{-\:\:\: -\:\:\:\:\:\:\:\:-\:\:\:\:\:\:}\\8S\:\:\:\:\:\:\:\:\:\:\:\:\:=40\\S\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=5$

The total number of short answers is 5.

Put S=5 in equation 1.

$5+M=20\\M=15$

Hence, the number of multiple-choice questions are 15.

3 0

4 years ago