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3 years ago

11

The local music activities coordinator sold 300 tickets to the orchestra concert. Student tickets were $4, and adult tickets wer

e &6. If the total sales were $1,600, how many student tickets were sold?

1 answer:

VARVARA [1.3K]3 years ago

4 0

There were sold 100 student tickets

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Describe two different ways that you could find the product 8 x 997 using mental math. Find the product and explain why your met

jolli1 [7]

One method is to round 997 up to 1000, multiply by 8, and then subtract 8 times 3. This would give you the solution of 7976.

Another method (Which I personally wouldn't use) is to recursively double 997. This is more difficult, although effective. After you double 997, double the resulting number, and then double the resulting number from that, you have the solution. This is because 2^3 is 8.

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2 years ago

The product of two decimal is 1.5008 if one of them is 0.56 find the other?<br>

babymother [125]

Answer:

divide 1.5008 and 0.56

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3 years ago

The total number of people at a football game was 5600. Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

rewona [7]

Answer:

1100 field-side tickets and 4500 end-zone tickets.

Step-by-step explanation:

Let x represent number of field side tickets and y represent number of end-zone tickets.

We have been given that the total number of people at a football game was 5600. We can represent this information in an equation as:

$x+y=5600...(1)$

$y=5600-x...(1)$

We are also told that Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

Cost of x field side tickets would be $40x$ and cost of y end-zone tickets would be $20y$ .

The total amount of money received for the tickets was $134000. We can represent this information in an equation as:

$40x+20y=134000...(2)$

Upon substituting equation (1) in equation (2), we will get:

$40x+20(5600-x)=134000$

$40x+112000-20x=134000$

$20x+112000=134000$

$20x+112000-112000=134000-112000$

$20x=22000$

$\frac{20x}{20}=\frac{22000}{20}$

$x=1100$

Therefore, 1100 field side tickets were sold.

Upon substituting $x=1100$ in equation (1), we will get:

$y=5600-1100$

$y=4500$

Therefore, 4500 end-zone tickets were sold.

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3 years ago

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in s

Jobisdone [24]

Answer:

$f(x)=x^4-9x^2-50x-150$

Step-by-step explanation:

Let f(x) be the polynomial function of minimum degree with real coefficients whose zeros are 5, -3, and -1 + 3i be f(x).

By the complex conjugate property of polynomials, -1-3i is also a root of this polynomial.

Therefore the polynomial in factored form is $f(x)=(x-5)(x+3)(x-(-1+3i))(x-(-1+3i))$

We expand to get: $f(x)=(x^2-2x-15)(x^2+2x+10)$

We expand further to get:\

$f(x)=x^4-9x^2-50x-150$

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3 years ago

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses.

Luden [163]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

If we had increased the confidence level, the margin of error also would have increased.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96. Increasing the confidence level, z also increases, hence the margin of error also would have increased.

The sample size and the estimate are given as follows:

$n = 450, \pi = 0.47$ .

The lower and the upper bound of the interval are given, respectively, by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161$

The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

More can be learned about the z-distribution at brainly.com/question/25890103

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1 year ago