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Readme [11.4K]
3 years ago
12

Two families stayed at the same hotel in identical rooms. The first family stayed for 5 nights and paid extra to use a rollaway

bed for 2 of the nights. Their total bill was $413. The second family stayed for 3 nights and used a rollaway bed for 1 night. Their total bill was $243.
Part A
Write a system of equations that can be used to find, h, the cost per night of renting a room at the hotel, and b, the cost per night of using a rollaway bed at the hotel.





Part B
Use the elimination method to solve the system of equations from Part A for h and b. Show your work and explain the steps you take.
Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Part A: The system of equations that can be used is

5h + 2b = 413

3h + b = 243

Part B: The cost per night of renting a room is $73 and the cost per night using a roll away bed is $24

Step-by-step explanation:

Two families stayed at the same hotel in identical rooms

  • The first family stayed for 5 nights and paid extra to use a roll away bed for 2 of the nights.
  • Their total bill was $413
  • The second family stayed for 3 nights and used a roll away bed for 1 night.
  • Their total bill was $243

We need to write a system of equations that can be used to find, h, the cost per night of renting a room at the hotel, and b, the cost per night of using a roll away bed at the hotel and find the values of h and b using the elimination method

Part A:

∵ The first family stayed for 5 nights

∵ The cost of the renting room per night is $h

∴ The cost of the five nights = 5h

∵ The first family paid extra to use a roll away bed for 2 of the nights

∵ The cost of the roll away bed per night is $b

∴ The cost of the two nights = 2b

∵ Their total bill was $413

∴ 5h + 2b = 413 ⇒ (1)

∵ The second family stayed for 3 nights

∵ The cost of the renting room per night is $h

∴ The cost of the five nights = 3h

∵ The first family paid extra to use a roll away bed for 1 of the night

∵ The cost of the roll away bed per night is $b

∴ The cost of the two nights = b

∵ Their total bill was $243

∴ 3h + b = 243 ⇒ (2)

The system of equations that can be used is

5h + 2b = 413

3h + b = 243

Bart B:

∵ 5h + 2b = 413 ⇒ (1)

∵ 3h + b = 243 ⇒ (2)

- Multiply equation (2) by -2 to eliminate b

∴ -6h - 2b = -486 ⇒ (3)

- Add equations (1) and (3)

∴ -h = -73

- Multiply both sides by -1

∴ h = 73

Substitute the value of h in equations (1) or (2) to find b

∵ 3(73) + b = 243

∴ 219 + b = 243

- Subtract 219 from both sides

∴ b = 24

The cost per night of renting a room is $73 and the cost per night using a roll away bed is $24

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#LearnwithBrainly

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Answer:

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Answer:

The probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

Step-by-step explanation:

Let a set be events that have occurred be denoted as:

S = {A₁, A₂, A₃,..., Aₙ}

The Bayes' theorem states that the conditional probability of an event, say <em>A</em>ₙ given that another event, say <em>X</em> has already occurred is given by:

P(A_{n}|X)=\frac{P(X|A_{n})P(A_{n})}{\sum\limits^{n}_{i=1}{P(X|A_{i})P(A_{i})}}

The disease Breast cancer is being studied among women of age 60s.

Denote the events as follows:

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The information provided is:

P(B) = 0.0312\\P(+|B)=0.81\\P(+|B^{c})=0.92

Compute the value of P (B|+) using the Bayes' theorem as follows:

P(B|+)=\frac{P(+|B)P(B)}{P(+|B)P(B)+P(+|B^{c})P(B^{c})}

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Thus, the probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

4 0
3 years ago
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HELP ASAP PLZZZZ
Tcecarenko [31]
QUESTION 1

The given system of equations is

3d - e = 7...eqn(1)
d + e = 5...eqn(2)

To solve by linear combination, we add equation (1) to equation (2) to get,

3d  + d= 7 + 5


4d = 12


We divide through by 4 to obtain,


d =  \frac{12}{4}


d = 3


We put d=3 into equation (2) to get,



3+ e = 5


e = 5 - 3


e = 2


\boxed {The \: solution \: is  \: (3, 2)}



QUESTION 2


The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)


To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

4x - 3x = 5 - 3



This implies that,

x = 2


Put x=3 into equation (1) to get,

4(2) + y = 5

8+ y = 5


y = 5 - 8



y =  - 3

The solution is

(2,-3)



QUESTION 3

We want to solve the system;


a – 2b = –2 ....eqn(1)


2a + 2b = 14...eqn(2)

by linear combination.


We need to add equation (1) to equation (2) to eliminate b.


This implies that,

2a + a = 14 +  - 2




Simplify,

3a = 12



Divide both sides by 3 to get,


a = 4
Put a=4 into equation (2) to obtain,



2(4) + 2b = 14


8 + 2b = 14
2b = 14 - 8


2b = 6


b = 3


The ordered pair in the form (a, b) is

(4,3)



QUESTION 4

The given system of equations is


11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)


We subtract equation (2) from equation (1) to get,


11x - 3x = 18 - 2


8x = 16


x = 2


Put x=2 into equation (2) to obtain,


3(2) + 4y = 2


This implies that,


6 + 4y = 2


4y = 2 - 6


4y =  - 4


y=-1

The correct answer is (2,-1).




QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2


We add the two equations to eliminate e.


This implies that,

2d + d = 8 + 4


3d = 12



We divide both sides by 3 to get,


d = 4


We put d=4 into equation (2) to get,

4 - e = 4

- e = 4 - 4



- e = 0



e = 0


The solution is

(4,0)
7 0
3 years ago
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