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vredina [299]
3 years ago
15

У

Mathematics
1 answer:
jasenka [17]3 years ago
5 0

Answer:

oops! it's too hard!!

i think i should need to do first so that i could provide you correct answer of it!!

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A total of 768 tickets were sold for the school play. They were either adult tickets or student tickets. There were 68 more
sergejj [24]

Answer:

Let x = the number of adult tickets sold

Let x + 61 = the number of student tickets sold

x + x + 61 = 761

2x = 700

x = 350

x + 61 = 411

8 0
3 years ago
Read 2 more answers
Najat is selling raffle tickets. The table shows the cost (in dollars) of buying different numbers of tickets.
oksano4ka [1.4K]

Answer:

The equation relating x and y is: y=4x

Step-by-step explanation:

Let x represent number of tickets

and y represent cost

a) Plot the ordered pairs

The ordered pairs (1,4),(2,8),(3,12),(4,16) is plotted in figure attached.

b)  Write an equation relating x and y

y= 4x

Now checking by putting values of x to see, if we get values of y

if x=1,  then y=4x => y=4*1 =>y=4

if x=2 then y=4x => y=4*2 =>y=8

if x=3 then y=4x => y=4*3 =>y=12

So, we get same values as table.

The equation relating x and y is: y=4x

5 0
3 years ago
Situation
Naily [24]

Answer:

egwegerhreherhsrhs

Step-by-step explanation:

thtrhrtsjsjhshjshhs

3 0
3 years ago
16. In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-
sergey [27]

Answer:

Step-by-step explanation:

Total number of students = 68

Let history = H

Maths = M

English = E

n(H) = 25

n(M) = 25

n(E) = 34

n(HnMnE) = 3

Total = n(H) + n(M) + n(E) - people in exactly two groups + 2(people in exactly 3 groups) + people in none of the groups

68 = 25 + 25 + 34 - people in exactly two groups - 6 +0

68 = 84 -6 - people in exactly two groups

68 = 78 - people in exactly two groups

People in exactly two groups = 78 - 68

= 10

OR

From the venn diagram, people in exactly two groups are represented by x, y and z

Total = 25 - x - y - 3 + 25 - x - z - 3 + 34 - y - z - 3 + x + y + z + 3

68 = 50 - x - 3 + 34 - y - z - 3

68 = 84 - 6 - x - y - z

68 = 78 - x - y - z

68 - 78 = - x - y - z

-10 = -(x + y + z)

x+y+z = -10/-1

x+y+z = 10

The number of students that registered for exactly two courses = 10

4 0
3 years ago
Read 2 more answers
Joe went to the fair and tried his hand at the shooting gallery. He earned 20 points each time he hit the target but lost 50 poi
Lubov Fominskaja [6]
Answer: Joe hit the target 4 times.

Explanation: We can write this scenario as a system of equations.

Let’s express the number of times he hits the target with x.
Let’s express the number of times he misses the target with y.

“He earned 20 points each time he hit the target but lost 50 points when he miss. Joe ended the night with negative 470 points...”

20x - 50y = -470

“...after 15 shots.”

x + y = 15

Let’s write the whole system of equations.

20x - 50y = -470
x + y = 15

Let’s solve the second equation for y.

x + y = 15

Subtract x from both sides.

y = 15 - x

Let’s substitute y in the first equation with 15 - x.

20x - 50(15 - x) = -470

Distribute -50 among 15 and -x in the term -50(15 - x).

20x - 750 + 50x = -470

Combine like terms on the left side.

70x - 750 = -470

Add 750 on both sides.

70x = 280

Divide both sides by 70.

x = 4

Since we know that x = 4, we know that Joe hit the target 4 times.

8 0
3 years ago
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