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goldfiish
2 months ago
12

For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How

much are lectures per hour?
Mathematics
2 answers:
zzz [12.3K]2 months ago
4 0
12 lectures in one hour

Inessa [12.5K]2 months ago
3 0

Answer

Lecture cost = $7.33 each hour

Explanation

Let e denote the expense for exam hours.

Let w signify the expense for workshop hours.

Let l indicate the expense for lecture hours.

The problem states that exam hours are priced at double that of workshop hours, represented by:

e=2w equation (1)

It also indicates that workshop hours are at twice the cost of lecture hours, exemplified by:

w=2l equation (2)

In addition, we understand that the combined cost for 3 hours of exams, 24 hours of workshops, and 12 hours of lectures totals $528, represented as:

3e+24w+12l=528 equation (1)

Now, let’s compute the value of l:

Step 1. Determine l via equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Substitute equation (1) into equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Insert equation (2) into equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Lecture cost = $7.33 every hour



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lawyer [12517]

Answer:

The 10th term in the geometric progression is 29.

Step-by-step explanation:

Given: In a geometric series, [T3 = 18] and [T6 = 486].

To find: The term [T10]?

Solution:

A geometric sequence takes the form [a, ar, ar^2,...]

Where, a represents the first term, and r denotes the common ratio.

The nth term is expressed as [Tn = a * r^(n-1)]

From the information provided: [T3 = a * r^2 = 18]

And [T6 = a * r^5 = 486]

By dividing the second equation by the first:

[(a * r^5) / (a * r^2)] = 486 / 18

[r^3 = 27]

Taking the cube root provides: r = 3.

Inserting r into one of the equations allows us to solve for a.

Substituting r gives: [T3 = a * r^2 = 18]

Thus, the first term is a = 2, and the common ratio is r = 3.

The 10th term in the geometric progression is computed as:

[T10 = a * r^(10-1)]

[Thus, T10 = 29.]

8 0
2 months ago
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On a website selling books, three paperback books and five hardcover books cost $80.10, and seven paperback books and four hardc
Zina [12379]
To solve this problem, you'll want to substitute the first equation into the second or the other way around. The equations given are: 1. 3 paperback books + 5 hardcover books = $80.10; 2. 7 paperback books + 4 hardcover books = $100.65. It is helpful to rearrange the first equation to find 5 hardcover books = $80.10 - 3 paperback books, leading to hardcover book = $16.02 - 0.6 paperback books. Now, substitute this into the second equation: 7 paperback books + 4 ($16.02 - 0.6 paperback books) = $100.65, which simplifies to 7 paperback books + $64.08 - 2.4 paperback books = $100.65. This results in 4.6 paperback books = $100.65 - $64.08 = $36.57, thus paperback book = $7.95. You can then use this price in the first equation to determine the hardcover book price: 3 paperback books + 5 hardcover books = $80.10, substituting gives 3($7.95) + 5 hardcover books = $80.10, which leads to 5 hardcover books = $80.10 - $23.85 = $56.25, therefore hardcover book = $11.25. Hence, the total cost for one paperback and one hardcover book is $7.95 + $11.25 = $19.20. 
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1 month ago
A journalist wants to determine the average annual salary of CEOs in the S&P 1,500. He does not have time to survey all 1,50
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Answer:

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A journalist intends to find the average annual salary of CEOs within the S&P 1,500. Due to time constraints, a sample is used.

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it’s $295.47 that’s everything

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