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Delicious77 [7]
3 years ago
13

Bonifacio works 40 hours per week as a customer service representative. If he made $17,680 last year, how much was he paid per h

our?
Mathematics
2 answers:
oee [108]3 years ago
7 0

Answer:

Bonifacio was paid $8.5 per hour.

Step-by-step explanation:

To know how much Bonifacio earned per hour, considering that at the end of the year he had made $ 17,680, and that he works 40 hours a week, we must first know how many weeks a year has.

A year has 52 weeks, therefore, Bonifacio earned $ 340 weekly (since 17,680 / 52 = 340). So, if I work 40 hours a week, for every hour of work Bonifacio made $ 8.5 (since 340/40 = 8.5).

GaryK [48]3 years ago
6 0
$8.50 because if he worked 40 hours a week you would multiple that by 52 because there are 52 weeks in a year. From there you would take the total he made and divide it by the hours he worked.
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Answer:

1    $16.50

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Step-by-step explanation:

Please find attached the complete question

Total cost in dollars = fixed cost + (variable cost x number of CDs bought)

fixed cost = $4.50

variable cost = $12

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total cost in dollars = $4.50 + $12c

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2 years ago
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Step-by-step explanation:

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3 years ago
What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
a_{n+1}=a_{n-3}+4d

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

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a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

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Well, since a_n is just the term after a_{n-1} scaled by r, you can write

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Doing this again and again, you'll see a similar pattern emerge:

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a_{n+1}=r^3(ra_{n-3})
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a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

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Answer:

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Step-by-step explanation:

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