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CaHeK987 [17]
2 years ago
5

A flu epidemic is spreading through a town of 48,000 people. It is found that if x and y denote the numbers of people sick and w

ell in a given week respectively.
And if s and w denote the corresponding numbers for the following week, then

1/3 x + ¼ y = s

2/3 x + ¾ y = w

1) write this system of equations in a matrix form

2)Suppose that 13,000 people are sick in a given week. how many were sick the preceding week?
Mathematics
1 answer:
Liula [17]2 years ago
8 0

Answer:

a) The simultaneous equation represented in matrix form, is

[1/3 1/4] [x] = [s]

[2/3 3/4] [y] = [w]

Ax = B

[1/3 1/4] = matrix A (matrix of coefficients)

[2/3 3/4]

[x] = matrix x (matrix of unknowns)

[y]

[s] = matrix B (matrix of answers)

[w]

b) Number of sick people the preceding week = 12005

Step-by-step explanation:

x = Number of sick people in a week

y = Number of people that are well in a week

s = Number of sick people the following week

w = Number of people that are well the following week.

The relationship between these is given as

(1/3)x + (1/4)y = s

(2/3)x + (3/4)y = w

In matrix form, this is simply presented as

[1/3 1/4] [x] = [s]

[2/3 3/4] [y] = [w]

which is more appropriately written as

Ax = B

where

[1/3 1/4] = matrix A (matrix of coefficients)

[2/3 3/4]

[x] = matrix x (matrix of unknowns)

[y]

[s] = matrix B (matrix of answers)

[w]

b) Taking the current conditions as s and w, then the preceding week will be x and y

The number of sick people in this week, s = 13000

The number of people well in this week, w = total population - Number of sick people.

w = 48000 - 13000 = 35000

So, the simultaneous equation becomes

(1/3)x + (1/4)y = 13000

(2/3)x + (3/4)y = 35000

Then we can solve for the number of sick and well people the preceding week.

We can solve normally or use matrix solution.

Ax = B

x, the matrix of unknowns is given by product of the inverse of A (inverse of the matrix of coefficients) and B (matrix of answers)

x = (A⁻¹)B

But, solving normally,

(1/3)x + (1/4)y = 13000

(2/3)x + (3/4)y = 35000

x = 12004.8 = 12005

y = 35995.2 = 35995

Number of sick people the preceding week = x = 12005

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1 year ago
Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0
2 years ago
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