The drama club was selling ticketsto the school play. Adult ticketscost $8.00 each, and studenttickets cost $5.00 each. The littletheater holds 142 people and wassold out for both Friday andSaturday. The total sales for thetwo days was $1,948.00.1. How many adult tickets weresold out over the two days?2. How many student tickets weresold out over the two days?

Answers

Answer 1

We are given a problem that can be solved using a system of linear equations. Let A, be the number of adults, and S the number of students. Since there are in total 142 people and there were two days, this means that the sum of the number of adults and the number of students must be 284, which can be written mathematically as follows:

[tex]A+S=284,(1)[/tex]

This is our first equation. The second equation is found using the total sales of $1948. Since the ticket per adult is $8 and per student is $5, we have the following equations:

[tex]8A+5S=1948,(2)[/tex]

To solve this equation we will solve for A in equation (1), by subtracting S to both sides;

[tex]\begin{gathered} A+S-S=284-S \\ A=284-S \end{gathered}[/tex]

Now we will replace this value in equation (2):

[tex]8(284-S)+5S=1948[/tex]

Now we will apply the distributive property:

[tex]2272-8S+5S=1948[/tex]

Addins like terms:

[tex]2272-3S=1948[/tex]

Subtracting 2272 to both sides;

[tex]\begin{gathered} 2272-2272-3S=1948-2272 \\ -3S=-324 \end{gathered}[/tex]

Dividing both sides by -3:

[tex]S=-\frac{324}{-3}=108[/tex]

Now we replace this value in equation (1), where we have already solved for A:

[tex]\begin{gathered} A=248-108 \\ A=140 \end{gathered}[/tex]

Therefore, there were sold 108 student tickets and 140 adult tickets.

Related Questions

Use the substitution u = (2x - 2) to evaluate the integral x³e(^2x^4-2) dx

Answers

The substitution u = (2x - 2) to the integral x³e(^2x^4-2) dx is (2x – 2)/4 +c

What is meant by integral?In mathematics, an integral assigns numerical values to functions in order to describe concepts like displacement, area, volume, and other outcomes of the combination of infinitesimally small data. Integral discovery is a process that is referred to as integration. One of the fundamental, crucial operations of calculus, along with differentiation, is integration[a]. It can be used to solve issues in mathematics and physics involving, among other things, the volume of a solid, the length of a curve, and the area of an arbitrary shape. The integrals listed here are those that fall under the category of definite integrals, which can be thought of as the signed area of the region in the plane that is enclosed by the graph of a particular function between two points on the real line.

Therefore,

Use the substitution

U = (2x -2)

to evaluate integral x³e(^2x^4-2) dx

let u = 2x -2

du = x dx or dx =du/2

u = 2x-2

du = d(2x – 2)

du = 2dx

dx = du/2

∫ (2x -2)dx = ∫u du/2

=1/2 ∫u du

= ½ u square /2 +c

= u square /4 +c

= (2x – 2)/4 +c

To learn more about integral, refer to:

https://brainly.com/question/27419605

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Suppose that $16,065 is invested at an interest rate of 6.6% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time t, in years. b) What is the balance after 1 year? 2 years?5 years? 10 years? c) What is the doubling time?

Answers

Okay, here we have this:

Considering the provided information we obtain the following:

Replacing in the Compound Interest formula we obtain the following:

[tex]\begin{gathered} A(t)=Pe^{rt} \\ A(t)=16065e^{0.066t} \end{gathered}[/tex]

After 1 year (t=1):

[tex]\begin{gathered} A(1)=16065e^{0.066(1)} \\ A(1)\approx17,161.06 \end{gathered}[/tex]

We obtain that after one year the balance is aproximately $17,161.06.

After 2 years (t=2):

[tex]\begin{gathered} A(2)=16065e^{0.066(2)} \\ A(2)=18331.90 \end{gathered}[/tex]

We obtain that after two years the balance is aproximately $18,331.90

After 5 years (t=5):

[tex]\begin{gathered} A(5)=16065e^{0.066(5)} \\ A(5)=$22,345.90$ \end{gathered}[/tex]

We obtain that after five years the balance is aproximately $22,345.90.

After 10 years (t=10):

[tex]\begin{gathered} A(10)=16065e^{0.066(10)} \\ A(10)=$31,082.44$ \end{gathered}[/tex]

We obtain that after ten years the balance is aproximately $31,082.44.

In this case the doubling time will be when she has double what she initially had, that is: $16,065*2=$32130, replacing in the formula:

[tex]32130=16065e^{0.066t}[/tex]

Let's solve for t:

[tex]\begin{gathered} 32130=16065e^{0.066t} \\ 16065e^{\mleft\{0.066t\mright\}}=32130 \\ \frac{16065e^{0.066t}}{16065}=\frac{32130}{16065} \\ e^{\mleft\{0.066t\mright\}}=2 \\ 0.066t=\ln \mleft(2\mright) \\ t=\frac{\ln\left(2\right)}{0.066} \\ t\approx10.502years \end{gathered}[/tex]

Finally we obtain that the doubling time is approximately 10.502 years or about 10 years 6 months.

The average score for games played in the NFL is 22 and the standard deviation is 9.3 points. 41 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution.

a. What is the distribution of ¯x x¯
? ¯xx¯ ~ N( , )
b. What is the distribution of ∑x ? ∑x ~ N ( , )
c. P( ¯x > 19.8214) =
d. Find the 60th percentile for the mean score for this sample size.
e. P(20.6214 < x¯< 23.2262) =
f. Q1 for the x¯distribution =
g. P( ∑x > 829.0774) =

For part c) and e), Is the assumption of normal necessary? NoYes

Answers

Using the normal distribution and the central limit theorem, it is found that:

a) The distribution is: x¯ ~ N(22, 1.45).

b) The distribution is: ∑x ~ N(902, 59.55).

c) P( ¯x > 19.8214) = 0.9332 = 93.32%.

d) The 60th percentile for the mean score for this sample size is of 22.37 points a game.

e) P(20.6214 < x¯< 23.2262) = 0.6312 = 63.12%.

f) Q1 for the x¯distribution = 21 points a game.

g) P( ∑x > 829.0774) = 0.8888 = 88.88%.

Assumption of normality is not necessary, as the sample sizes are greater than 30.

Normal Probability Distribution

The z-score of a measure X of a variable that has mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score measures how many standard deviations the measure X is above or below the mean of the distribution, depending if the z-score is positive or negative.From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X in the distribution.By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].Also by the Central Limit Theorem, for the sum of n instances of a variable, the mean is of [tex]\n\mu[/tex] and the standard deviation is of [tex]\sigma\sqrt{n}[/tex].Finally, by the Central Limit Theorem, assumption of normality is only necessary when the sample size is less than 30.

For a single game, the mean and the standard deviation of the number of points scored are given as follows:

[tex]\mu = 22, \sigma = 9.3[/tex]

For the average of 41 games, the standard error is:

[tex]s = \frac{9.3}{\sqrt{41}} = 1.45[/tex]

Hence the distribution is: x¯ ~ N(22, 1.45).

For the sum of the 41 games, the mean and the standard error are given as follows:

41 x 22 = 902.[tex]s = 9.3\sqrt{41} = 59.55[/tex].

Hence the distribution is: ∑x ~ N(902, 59.55).

In item c, the probability is one subtracted by the p-value of Z when X = 19.8214, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem: