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Digiron [165]
4 years ago
10

Visitors to a carnival can buy an unlimited-ride pass for $50 or an entrance-only pass for $20. In one day, 282 passes were sold

for a total of $10,680. The following system of equations models this scenario:
50x + 20y = 10,680
x + y = 282

How many unlimited-ride passes were sold?
Mathematics
2 answers:
baherus [9]4 years ago
6 0

Answer:

The  unlimited-ride passes sold are equal to 168

Step-by-step explanation:

According to given scenario:

x = unlimited-ride passes

y = entrance-only pass

Given that:

50x + 20y = 10,680  -------- eq1

x + y = 282  ------------ eq2

From eq2:

x = 282 - y

Putting value of x  in eq1:

50(282 - y) + 20y = 10,680

By simplifying:

14,100 - 50y + 20y = 10,680

14,100 - 30y = 10,680

30y = 14,100 - 10680

30y = 3,480

Dividing both sides by 30

y = 114

Now put value of y in eq2:

x + 114 = 282

x = 282 - 144

x = 168

So, the  unlimited-ride passes sold are equal to 168

i hope it will help you!

neonofarm [45]4 years ago
5 0

Answer:

option c

Step-by-step explanation:

The  unlimited-ride passes sold are equal to 168 so therefor it is c

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To pass science, a student must earn at least a grade of 70. How many students failed this science class?​
Colt1911 [192]

Answer:

7 students.

Step-by-step explanation:

To pass science, a student requires at least 70 grades so students who scored below 70 will be failed.

From the given bar chart we can calculate the number of students who scored below 70 grades

Students who earned 50 - 59 = 2

Students who earned 60 - 69 = 5

Therefore, number of students who failed were

2 + 5 = 7 students.

4 0
4 years ago
Read 2 more answers
How to answer this? A farmer had 455 ducks and chickens. He sold 2/5 of the ducks and bought another 104 chickens. As a result,
julia-pushkina [17]

Answer:

(a) 25

(b) $4816

Step-by-step explanation:

Let's say the starting amount of ducks is d, and the starting amount of chickens is c. Since there are only ducks and chickens, we can say that d + c= 455.

Next, the farmer sells 2/5 of the ducks, so we subtract 2/5 of the ducks so that the ending duck amount is d - (2/5)d = (3/5) d

After that, the farmer buys another 104 chickens, so we add 104 chickens to the current amount of chickens to get the ending chicken number as 104+c

Given the ending duck and chicken count, we can say that the ending farm animal count (we can represent this as a) is (3/5)d + 104 + c = a

The number of ducks is 2/3 the total number of animals at the end, so (3/5)d = (2/3)a

Let's list our equations out so it is easier to solve:

d+c = 455

(3/5)d + 104 + c = a

(3/5)d = (2/3)a

We have three equations and need to solve for three variables. One common variable in all three equations is d, so it might help if we put everything in terms of d.

Starting with d+c=455, if we subtract d from both sides, we get c=455-d. We can substitute this in the second equation to get

(3/5)d + 104 + 455 - d = a

-(2/5)d + 559 = a

(3/5)d = (2/3)a

Next, we can substitute for a. If we multiply both sides by 3 and then divide by 2 in (3/5)d = (2/3)a , we get

(9/10)d = a

Substitute that in to the second equation to get

-(2/5)d + 559 = (9/10) d

-(4/10)d + 559 = (9/10) d

add (4/10)d to both sides to isolate the variable and its coefficient

(13/10)d = 559

multiply both sides by (10/13) to isolate the coefficient

d = 430

Therefore, the starting number of ducks is 430 and the starting amount of chickens is 455-430 = 25

For (b), 2/5ths of the ducks are sold, so this is (430) * (2/5) = 172. Each duck is sold for 28 dollars, so this is 172*28=$4816 as the total price

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You have the option of buying 20 tickets at the fair for 15$ or 45 for 27$ which is te best deal
Lostsunrise [7]
45 tickets for 27$is definitely the best deal because if you go with the first option and buy it twice you will pay 25$for only 40 I can’t Explain properly but definitely 45 for 26$
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