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> 6th grade , Ratios,
& percentages , Intro
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The following table shows the speeds at which Jeff does 4 activities. For example, biking has a rate of 19
miles in 2 hours
Activity
Miles
Hours
19
2
Intro to rates
Biking
Walking
14
4
Solving unit rate problem
Running
15
3
Swimming
18
4
Solving unit price problem
Which type of activity does Jeff do at a speed of 4.5 miles per hour?
Practice: Unitrates
Choose 1
Rate problems
Biking
Practice: Rate problems
* Walking
> Comparing rates example
© Running
→ Practice: Comparing rates
Rate review
0 Swimming

"it's not the heat, it's the humidity".   that's a partly valid phrase you may have heard in the summer, but it's actually both.   the heat index, also known as the apparent temperature, is what the temperature feels like to the human body when relative humidity is combined with the air temperature.   this has important considerations for the human body's comfort.   when the body gets too hot, it begins to perspire or sweat to cool itself off.   if the perspiration is not able to evaporate, the body cannot regulate its temperature.   evaporation is a cooling process.   when perspiration is evaporated off the body, it effectively reduces the body's temperature.   when the atmospheric moisture content (i.e. relative humidity) is high, the rate of perspiration from the body decreases.   in other words, the human body feels warmer in humid conditions.   the opposite is true when the relative humidity decreases because the rate of perspiration increases.   the body actually feels cooler in arid conditions.   there is direct relationship between the air temperature and relative humidity and the heat index, meaning as the air temperature and relative humidity increase (decrease), the heat index increases (decreases).

step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

beluga wheeeaaalle matt

step-by-step explanation: