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3 years ago

What type of quadrilateral is DEFG? Select all that apply.

Mathematics

1 answer:

Ierofanga [76]3 years ago

5 0

Answer:

B. parallelogram

Step-by-step explanation:

Given

See attachment for quadrilateral

Required

Select the type of quadrilateral

From the attached image, we have:

$DE = FG$ -- Opposite, Parallel and Equal

$EF = GD$ -- Opposite, Parallel and Equal

Since all 4 sides are not equal, then it is neither a square; nor a rhombus

<em>Hence, the shape is a parallelogram</em>

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The sequence a1, a2, a3, . . . , an of n integers is such that ak = k if k is odd and ak = –ak – 1 if k is even. Is the sum of t

nasty-shy [4]

Answer:

Step-by-step explanation:

The

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3 years ago

63 plus what gives you 83

iVinArrow [24]

83-63=20
So your answer is 20

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3 years ago

1/3 of students at a school are boys. If there are 600 students at the school, how many are girls?

Temka [501]

Answer:

200 students are boys and rest of 400 are girls

Step-by-step explanation:

Given that:

Total students = 300

1/3 students of total are boys

so.

Boys = 1/3(600)

Boys = 200

For finding number of girls:

Girls = total - boys

Girls = 600 - 200

Girls = 400

i hope it will help you!

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3 years ago

Read 2 more answers

What is greater than 0.05 but less than 0.06

ArbitrLikvidat [17]

Something between 0.05 and between 0.06 so between those to could be 0.051 0.052 0.053 0.054 0.055 0.056 0.057 0.058 0.059

6 0

3 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

One person, thus $q = 1$ .

The expected value is:

$E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4$

The variance is:

$V = \frac{0.8}{0.04} = 20$

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0

1 year ago