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Lesechka [4]
3 years ago
6

any inequality w greater than or equal 3.2 represents the weight of each pumpkin in pounds that is allowed to be picked to be so

ld are listed. how many pumpkins can be sold? which pumpkins can be sold?
Mathematics
2 answers:
mel-nik [20]3 years ago
5 0
I dont see the list of the pumpkins that is allowed to be picked.
so i cant answer the question sorry
Kobotan [32]3 years ago
3 0
I don't see a list of pumpkins that can be sold. Can't help you. Sorry :(
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Find the perimeter of a quadrilateral with vertices at C (-1, 2), D (-2, -1), E (2, -2), and F (1, 1). Round your answer to the
babymother [125]

Answer:

B 12.68

Step-by-step explanation:

you will need to find the distance between each segment-

so the length of DC, CF, FE, ED

d=\sqrt{(x_{2}-x_{1})  ^{2}+(y_{2}-y_{1})  ^{2}  }

DC:

D(-2,-1) C(-1,2)

d=\sqrt{(-1--2)^{2} +(2--1)^{2} } \\d=\sqrt{1^{2}+3^{2}  } \\d=\sqrt{10} =3.16

CF:

C(-1,2) F(1,1)

d=\sqrt{(1--1)^{2} +(1-2)^{2} } \\d=\sqrt{2^{2}+-1^{2}  } \\d=\sqrt{4+1} \\d=\sqrt{x} 5=2.24

FE

F(1,1) E(2,-2)

d=\sqrt{(2-1)^{2} +(-2-1)^{2} } \\d=\sqrt{1^{2} +(-3)^{2} } \\d=\sqrt{1+9} \\d=\sqrt{10}= 3.16

ED

E(2,-2) D(-2,-1)

d=\sqrt{(-2-2)^{2} +(-1--2)^{2} } \\d=\sqrt{(-4)^{2}+1^{2}  } \\d=\sqrt{16+1} \\d=\sqrt{17}= 4.12\\

3.16+2.24+3.16+4.12= 12.68

4 0
3 years ago
E
Nata [24]

Answer:

Huh

Step-by-step explanation:

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
2 years ago
Original: $110<br> New: $59<br> %Decrease:__________
Lemur [1.5K]

Answer:

46.36%

Step-by-step explanation:

Original: 110

New: 59

Percent Decrease: x

110 - 59 = 51

51/110 ≈ 46.3636... ≈ 46.36%

-Chetan K

3 0
2 years ago
Read 2 more answers
Helps pls :-/<br> number 12<br> any help is greatly appreciated!
Alexandra [31]
X = 5, that’s the answer to question 12
6 0
3 years ago
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