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Nikolay [14]
3 years ago
6

The ballpark made a total of $15,000 from ticket sales at Wednesday's game. The ballpark charges $20 for each adult ticket and $

10 for each child's ticket. They sold 3 times as many children's tickets as adult tickets. Write a system of equations that can be used to determine the number of adult and child tickets sold. How many adult and child tickets were sold?
Mathematics
1 answer:
IRISSAK [1]3 years ago
8 0
<h3>Answer:</h3>

equations

  • 20a +10c = 15000
  • c = 3a

solution

  • 300 adult
  • 900 children's
<h3>Step-by-step explanation:</h3>

Let "a" and "c" represent the numbers of adult and children's tickets sold, respectively. The problem statement tells us two relationships between these values:

... 20a +10c = 15000 . . . . . . total revenue from ticket sales

... c = 3a . . . . . . . . . . . . . . . . relationship between numbers of tickets sold

Using the expression for c, we can substitute into the first equation to get ...

... 20a +10(3a) = 15000

... 50a = 15000

... a = 15000/50 = 300 . . . . . adult tickets sold

... c = 3·300 = 900 . . . . . children's tickets sold

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B.

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Hey there! I'm happy to help!

-----------------------------------------------------------------------------------------

<u>QUESTION 18</u>

Here is how you multiply a 2×2 matrix by a column vector (just one column matrix).

\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] *\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}ax&by\\cx&dy\\\end{array}\right]

Once you get that final matrix, you add the rows which will give you a final column vector. Let's do it with this question, as we are multiplying together the matrices A and v.

\left[\begin{array}{ccc}-1&4\\2&0\\\end{array}\right] *\left[\begin{array}{ccc}8\\6\\\end{array}\right] =\left[\begin{array}{ccc}-8&24\\16&0\\\end{array}\right] =\left[\begin{array}{ccc}16\\16\\\end{array}\right]

So, your answer is \left[\begin{array}{ccc}16\\16\\\end{array}\right].

-----------------------------------------------------------------------------------------

<u>QUESTION 14</u>

We have this equation of an ellipse and we need to solve for y. When we have a more complex equation to solve you get rid of the things furthest from the variable you are solving for first and then you get closer and closer to it.

\frac{x^2}{25} + \frac{y^2}{16}=1

We subtract \frac{x^2}{25} from both sides.

\frac{y^2}{16}=1-\frac{x^2}{25}

We multiply both sides by 16.

y^2=1-\frac{x^2}{25}*16

We square root both sides.

y= \sqrt{1-\frac{x^2}{25}*16

We can take a four out of our root (because 16 is a perfect square so it can be simplified more).

y= 4\sqrt{1-\frac{x^2}{25}

And we take out 1/5 (which is the square root of 1/25).

y=± \frac{4}{5}\sqrt{1-x^2

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<u>QUESTION 17</u>

We see that Sea Slugs have 3/10 matches won, so they have won 30% of their matches.

From now on, they will be winning 2/3 of their games. Let's look at the answer options.

A. If they play 15 more games, they will win 10 of them. Our total matches will be 25 (our original 10+15). So, we will have 13/25, which is equal to 52%, so this is incorrect.

B. If they play 10 more games they win 6 2/3 games (like that makes any sense) for a total of 9 2/3 games out of 20. This is about 48%, so this is incorrect.

C. If they play 40 games, they are winning fractional games again. They win 29 2/3 games out of 50 total, which is about 59%, but not quite 60. This is also incorrect.

D. If they play 45 matches, they will win 30 games. Add the original 10 and you have 40 games. Our total is 55. 40/55 is about 73%, so they have passed that 60% threshold, so this is the correct answer.

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<u>QUESTION 16</u>

Our question tells us that if we have two numbers x and y, and we add them, it has to be 24 or greater. By saying it can be equal to 24 or greater, we are going to be using a ≤ or ≥ sign. The opening of the arrow always faces the equal or greater quantity, and in this case it is x+y, so it will look like x+y≥24.

This eliminates answer choices B and D as they do not have the equal part on the inequality.

Product means to multiply, so it will be written as xy. It has to be less than 60, so in this case the greater quantity is 60, so the arrow will opening towards 60, so it will look like xy<60.

So, our final answer is A.

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<u>QUESTION 12</u>

You can only multiply matrices if the number of columns in the first matrix is equal to the number of rows in the second. So basically the two interior numbers have to be the same (a,b) and (b,c). So, if we follow this, we get these answers:

2×2 times 2×1 YES

2×1 times 2×2= NO

3×3 times 3×1= YES

3×1 times 3×3= NO

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Have a wonderful day and keep on learning! :D

6 0
2 years ago
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