Members of a softball team raised $2039.50 to go to a tournament. They rented a bus for $1157.50 and budgeted $49 per player for meals. Write and solve an equation which can be used to determine xx, the number of players the team can bring to the tournament.

Answers

Answer 1

Answer:

Answer:

18 = number of players

Step-by-step explanation:

Giving the following information:

Members of a softball team raised $2039.50 to go to a tournament. They rented a bus for $1157.50 and budgeted $49 per player for meals.

To calculate the total number of players they can bring, we need to use the following formula:

Total amount of money= fixed cost + unitary variable cost*number of players

2,039.5= 1,157.5 + 49*number of players

882/49= number of players

18 = number of players

Answer 2

Answer:

The team can bring a maximum of 18 players to the tournament given the amount they raised and the budgeted expenses.

Let's use "x" to represent the number of players the team can bring to the tournament.

The total amount raised by the team is $2039.50,

and they rented a bus for $1157.50. Each player's meal will cost $49.

The total amount spent on the bus and meals for x players can be represented as follows:

Total Expenses = Bus Cost + (Number of Players) * (Cost per Player's Meal)

= $1157.50 + x * $49

Since the team's total expenses should not exceed the total amount raised,

$2039.50 ≥ $1157.50 + x * $49

$2039.50 - $1157.50 ≥ x * $49

$882 ≥ x * $49

Now, divide both sides by $49 to solve for x:

x ≤ $882 / $49

x ≤ 18

So, the team can bring a maximum of 18 players.

Learn more about Inequality here:

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One of the factors of quadratic is (x+5) the only other piece of information is that the middle term is zero what must the other factor be

Answers

Answer:

x-5

Step-by-step explanation:

The middle term has a coefficient that is the sum of the constants in the binomial factors. For that to be zero, the other binomial factor must be ...

(x -5)

(x +5)(x -5) = x^2 -25 . . . . . middle term is zero

Given input 3823, 8806, 8783, 2850, 3593, 8479, 1941, 4290, 8818, 7413, and a hash function h(x = x mod 13, show the resulting separate chaining table.

Answers

Hello:
3823=13(294)+1
, 8806=13(677)+5
8783=13(675)+8
2850=13(219)+3
3593=13(276)+5
8479=13(652)+3
1941=13(149)+4
4290=13(330)
8818=13(678)+4
7413=13(570)+3

3823=1 mod 13
8806 = 5mod 13
8783= 8 mod 13 ........continue

Someone please answer!

Answers

Answer:

m=1/2

Step-by-step explanation:

y1= 1

y2=6

x1= -10

x2-0

m= slope

m= y2-y1/x2-x1

m=6-1/0 - - 10

m= 6-1/0+10

m=5/10

m=1/2

How to solve this w2-64= by facturing

Answers

Answer: (x+8)(x-8)

Step-by-step explanation:

hope this helped :)

among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 played both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball

Answers

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation of the answer:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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The table below shows the linear function f.x: -6,-5,-4,-3,-2
f(x): 22,19,16,13,10
Determine the average rate of change of the given function over an interval [-5, -2].
A. -1/3
b.1/3
c.-3
D.3