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rewona [7]
3 years ago
8

A mechanic charge $75 for up to 1 hour of work.for each additional hour,he charges $25 .

Mathematics
2 answers:
Maurinko [17]3 years ago
5 0

He charges 100 dollars for two hours?

sergiy2304 [10]3 years ago
3 0

He charged $75 for an hour. He then charges $25 for additional hours.

The equation for that would be: 75 + 25h = x

x = the final price. h = every our after the first hour.

So, for 1 hour he charges $75.

For 2 hours he charges $100.

For 3 hours he charges $125.

For 4 hours he charges $150.

For 5 hours he charges $175. And so on.

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12-packs of Skwunch apple juice cost $3.00 after a 40% discount. Which equation would allow you to find the original price of th
GaryK [48]

Let x be the original price.

Discount percentage = 40%

Price after discount = $3

So, x - \frac{40}{100} x = 3

\frac{100x-40x}{100} = 3

\frac{60x}{100} = 3

x = 3(\frac{100}{60} )

x=\frac{100}{20}

x = 5

Hence, the equation that would allow us to find the original price is 0.6x = 3.

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3 years ago
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what are the next 6 terms 10,14,18? and what pattern does it show? increasing decreasing? please Help. :(
Gnoma [55]

It looks like the next six terms are 22, 26, 30, 34, 38, 42

It is increasing by 4

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William is drafting his fantasy basketball team. He needs to select one player for each position. The following table shows how
denpristay [2]
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In an experiment, a fair coin is tossed 13 times and the face that appears (H for head or T for tail) for each toss is recorded.
zalisa [80]

Answer:

1) 1 element

2) 13 elements

3) 22 elements

4) 40 elements

Step-by-step explanation:

1) Only one element will have no tails: the event that all the coins are heads.

2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.

3) There are {13 \choose 2} = 78 elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are {11 \choose 2} = 55 possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22

4) We can have:

  • A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
  • One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)

For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)

For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).

This gives us a total of 18+22 = 40 possibilities.

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3 years ago
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