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weqwewe [10]
2 years ago
6

Middletown sponsors a two day conference for selected middle school students to study government. Suppose 20 student delegates w

ill the attend the conference Each school should be represented fairly in relation to it's population.
Mathematics
1 answer:
grandymaker [24]2 years ago
5 0

Answer:

  • North Middle School = 10 students
  • Central Middle School = 6 students
  • South Middle School = 4 students

Step-by-step explanation:

First calculate the total number of students in all the schools:

= 618 + 378 + 204  

= 1,200 students

Use this number to calculate the proportion of the total population that a school so that this can then be used to determine the proportion of the 20 delegates they should sent.

North Middle school:  

= 618/1,200 * 20  

= 10 students

 

Central Middle School:  

= 378 / 1,200 * 20  

= 6 students

 

South Middle School:  

= 204/1,200 * 20  

= 4 students  

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astraxan [27]
Divide the new length M'P' over the old one MP

(M'P')/(MP) = 4/10 = 0.4

The scale factor is 0.4

Because the scale factor is less than 1, it indicates the image (aka new figure) is smaller than the preimage (aka original figure). Basically the figure has been scaled down.
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Blababa [14]
First, answer the problem in the parentheses. 2 + 3 is 5. 5 x 4 is 20. Now, we are going to find the question that has an answer of 20. A = 11 so it cant be that one. B is 9 so not that one, either. C is 14 so not that one, but D = 20 so we have found the answer. D is the correct answer.
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The side length of a square is represented by the expression n-15
erik [133]
91819 n-19 +12=092!o
8 0
3 years ago
Five independent flips of a fair coin are made. Find the probability that(a) the first three flips are the same;(b) either the f
Korvikt [17]

Answer:

The answers to the question are

(a) 1/4

(b) 7/16

(c) 1/16

Step-by-step explanation:

To solve the question we note that

Total number  of outcomes = 32

Probability of the event of first three flips are the same =P(F)

Probability of the event of last three flips are the same =P(L)

Total number of outcomes = 2⁵ = 32

The number of ways in which the frist three flips are the same is

F  TTTTT, TTTTH, TTTHH, HHHTT, HHHHT, HHHHH, TTTHT, HHHTH = 8

L:  TTTTT, HHTTT, HTTTT, THTTT, HHHHH, TTHHH, HTHHH, THHHH = 8

The probability that the first and the last three flips are the same that is

F ∩ L; TTTTT, HHHHH = 2

Therefore P(F ∩ L) = 2/32

(a) P(F) = 8/32 =1/4 also

     P(L) = 8/32 =1/4

(b) P(LUF) =  P(L) + P(F) - P(F ∩ L) = 1/4+1/4-1/16 =7/16

(c) Let the event of at least two heads among the first three flips be H

and the event of at least two tails among the last three flips be T

Then we have

H; HHHHH, HHHHT, HHHTT, HHHTH, HHTTT, HHTHH, THHTT, THHHT

= 8

T; TTTTT. HTTTT, HHTTT, THTTT, THHTT, HHHTT, THHTT, HTTTH =8

Also H∩T  = TTTHH, HHTTT = 2

Therefore P(H ∩ T) = 2/32 = 1/16

5 0
2 years ago
Let x ∼ bin(9, 0.4). find
Kobotan [32]
A.
\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)
=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}
\approx0.025

b.
\mathbb P(X\ge2)=1-\mathbb P(X
=1-\dbinom90(0.4)^0(1-0.4)^{9-0}-\dbinom91(0.4)^1(1-0.4)^{9-1}
\approx0.929

c.
\mathbb P(2\le X
=\dbinom92(0.4)^2(1-0.4)^{9-2}+\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}
\approx0.663

d.
\mathbb P(2
=\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}+\dbinom95(0.4)^5(1-0.4)^{9-5}
\approx0.669

e.
\mathbb P(X=0)=\dbinom90(0.4)^0(1-0.4)^{9-0}\approx0.01

f.
\mathbb P(X=7)=\dbinom97(0.4)^7(1-0.4)^{9-7}\approx0.021

g, h.
For X\sim\mathcal B(n,p), recall that \mathbb &#10;E[X]=\mu_X=np and \mathbb V[X]={\sigma_X}^2=np(1-p). So

\mu_X=9(0.4)=3.6
{\sigma_X}^2=9(0.4)(0.6)=2.16
6 0
3 years ago
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