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uranmaximum [27]

2 years ago

15

Janie ordered boxed lunches for a student advisory committee meeting. Each lunch costs $4.25. The total cost of the lunches is $

53.75, including a $7 delivery fee.

2 answers:

andriy [413]2 years ago

7 0

Answer:

11 boxed lunches

Step-by-step explanation:

Full question

Janie ordered boxed lunches for a student advisory committee meeting. Each lunch cost 4.25. The total cost of the lunches is 53.75, including a 7$ delivery fee. Write and solve an equation to find x the number of boxed lunches Janie ordered

First of all subtract the delivery feesince it was inckuded in the total cost, this will now be the total cost of all the noxed lunches ordered by Janie, then divide the balance of the total cost by the cost of one boxed lunch to get thd total boxed kunches

X= 53.75-7/4.25

X= 53.75-7= 46.75/4.25

X=11

Shkiper50 [21]2 years ago

7 0

Answer:

The total number of lunch boxes is 11.

<em>⇒ 4.25x + 7 = 53.75 (equation)</em>

<u>Step-by-step explanation:</u>

If 1 lunch box costs $ 4.25

x lunch boxes costs x × 4.25 = 4.25x

Janie paid a total of $ 53.75 including delivery

<em>⇒ 4.25x + 7 = 53.75 (equation)</em>

4.25x = 53.75 - 7

4.25x = 46.75

∴x = $\frac{46.75}{4.25}$ = 11

<u>Therefore, Janie bought 11 boxes.</u>

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c) $\{ x \in C: M(x) = 1 \} = C$

d) $\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

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Step-by-step explanation:

M(x) = 1 if the person x came to the meeting, and 0 otherwise.

O(x) = 1 if the person is an officer of the club and 0 otherwise.

D(x) = 1 if the person has paid hid/her club dues and 0 otherwise.

Lets also call C the set given by the members of the club. C is the domain of the functions M, O and D.

a) If someone is not an officer, the there should be at least one value x such that O(x) = 0. This can be expressed by logic expressions this way

$\exists \, x \in C : O(x) = 0$

b) If all the officers came on time to the meeting, then for a value x such that O(x) = 1, we also have that M(x) = 1. Thus, the set of officers of the Club is contained on the set of persons which came to the meeting on time, this can be written mathematically this way:

$\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) If everyone was in time for the meeting, then C is equal to the set of persons who came to the meeting on time, or, equivalently, the values x such that M(x) = 1. We can write that this way:

$\{ x \in C: M(x) = 1 \} = C$

d) If everyone paid their dues or came on time to the meeting, then if we take the set of persons who came to the meeting on time and the set of the persons who paid their dues, then the union of the two sets should be the entire domain C, because otherwise there should be a person that didnt pay nor was it on time. This can be expressed logically this way:

$\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) If at least one person paid their dues on time and came on time to the meeting, then there should be a value x on C such that M(x) and D(x) are both equal to 1. Therefore

$\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) If there is an officer who did not come on time for the meeting, then there should be a value x in C such that O(x) = 1 (x is an officer), and M(x) = 0. As a result, we have

$\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

I hope that works for you!

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