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Setler [38]
3 years ago
6

A business buys invoice forms at a cost of $4.45 a box for the first 20 boxes, $4.00 a box for the next 25 boxes, and $3.75 a bo

x for any additional boxes. How many boxes of invoice forms can be bought for $234.00?
Mathematics
1 answer:
Sergio039 [100]3 years ago
3 0
So the first 20 boxes at 4.45 a piece will total 89.
The next 25 at 4 each will be 100.
Add those and you have 189 dollars, which is a spare of 45.
45/3.75 = 12 . 
So the answer is 57
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127 is a number, write it as a sum of 100s 10s and 1s
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Add the hundreds with the tens and the ones. 10+20+7=127
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Suppose you are representing the employees at a large corporation during contract negotiations. You have a list of the salaries
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Answer:

Median

Step-by-step explanation:

Median is ascertained by arranging a set of data in either ascending or descending order and picking the number that appears In the middle. The measure of center to use for this test is the median because it will represent the employees, it helps to arrange their income accordingly and the central item is determined.

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The average annual salary of the employees of a company in the year 2005 was $70,000. It increased by the same factor each year
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We can make use of the general formula for the geometric series to generate the function representing the average annual salary.
an = a0(r)^(n-1)

Or
f(x) = a0(r)^(x - 1)

Plugging in the given values for the year 2005 and 2006 to ge the value of r.
82000 = 70000 (r)^(1-1)
r = 1.1714

Therefore, the function is:
f(x) = 70,000 (1.1714)^(x-1)
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3 years ago
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Jack bought 3 protein bars for a total of $4.26 which equation could be used to find the cost c on dollars of each protein bar
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Answer: (C)

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3 0
3 years ago
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
Sedaia [141]

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

#SPJ4

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2 years ago
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