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Klio2033 [76]
3 years ago
13

A student wrote the equation 4(x+5.5)=3(1.5x+4.75) to represent the problem shown. Find her mistake and correct it.

Mathematics
1 answer:
Afina-wow [57]3 years ago
8 0

Answer:

  • $8.5

Step-by-step explanation:

<u>The first group consists of Petra and 4 friends, so the total number of members is 5, therefore the left side of equation should be:</u>

  • 5(x + 5.5)

<u>The second group consists of 4 people, Valentina and 3 friends, therefore the right side of the equation should be</u>:

  • 4(1.5x + 4.75)

<u>Solving the equation to find the cost of a ticket for Petra's group:</u>

  • 5(x + 5.5) = 4(1.5x + 4.75)
  • 5x + 27.5 = 6x + 19
  • 6x - 6x = 27.5 - 19
  • x = 8.5

Cost of a ticket is $8.5

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Step-by-step explanation:

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

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and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

(b) the required probability is given by considering X=0,1 and Y =0,1

f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

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Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

= 0.12

(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3

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