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kompoz [17]
3 years ago
13

In 2011, the average daily temperature in Darrtown was 65°F. In 2012, the average daily temperature increased by 3% but then dec

reased by 4.5% in 2013.
What was the daily average temperature in Darrtown in 2013?
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
8 0
You have been asking a lot to do with percents so I started working at leaning how to do them
65=100%
65/100=0.65
0.65=1%
3%=1.95
4.5%=2.925
65+1.95=66.95
66.95-2.925=64.025

64.025°F
I think I did it right
Sav [38]3 years ago
4 0

Answer:

<em>The daily average temperature in Darrtown in 2013 was 63.93725 °F</em>

Step-by-step explanation:

The average daily temperature in 2011 was  65° F.

In 2012, the temperature increased by 3%. So, <u>the amount of increased temperature</u> =(65\times 3\%)= (65\times 0.03)= 1.95 °F

So, <u>the temperature in 2012</u> =(65+1.95)= 66.95 °F

In 2013, the temperature is decreased by 4.5%. So, <u>the amount of decreased temperature</u> =(66.95\times 4.5\%)= (66.95\times 0.045)=3.01275 °F

So, the daily average temperature in 2013 =(66.95-3.01275)=63.93725 °F

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Answer:

The answer is: y = 2/3x - 3

Step-by-step explanation:

Given point: (3, -1)

Given equation: y = 2/3x - 5, which is in the form y = mx + b where m is the slope and b is the y intercept.

Parallel lines have the same slope. Use the point slope form of the equation with the point (3, -1) and substitute:

y - y1 = m(x - x1)

y - (-1) = 2/3(x - 3)

y + 1 = 2/3x - 6/3

y + 1 = 2/3x - 2

y = 2/3x - 3

Proof:

f(3) = 2/3(3) - 3

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andrew-mc [135]

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