0.805 mph equals 36 centimeters/seconds
Let's look at work rates per minute.
Together they can paint the wall in x minutes.
Working together, in 1 minute, they do 1/x of the job.
The student who paints the wall in 16 minutes does 1/16 of the job in 1 minute.
The student who paints the wall in 24 minutes does 1/24 of the job in 1 minute.
Together, they do 1/16 + 1/24 of the job in 1 minute, but from above, we see that together, they do 1/x of the job in 1 minute, so 1/16 + 1/24 must equal 1/x. That gives us our equation.
1/16 + 1/24 = 1/x
1/16 * 3/3 + 1/24 * 2/2 = 1/x
3/48 + 2/48 = 1/x
5/48 = 1/x
x = 48/5 = 9.6
Answer: It takes them 9.6 minutes, or about 10 minutes to do the job together.
Since the area of the poster doesn't change by putting it in a frame, we presume the question is asking what the area of the framed poster is.
The length of the poster in its frame is ...
(frame width on one side) + (poster length) + (frame width on the other side)
2 in + 32 in + 2 in = 36 in
Likewise, the width of the poster in its frame is ...
2 in + 24 in + 2 in = 28 in
The area of a rectangle 36 in by 28 in is the product of these dimensions:
Area = (36 in)×(28 in) = (36×28) in² = 1008 in²
Alright, so you have the basic formula- good.
You have the A value (400), the interest rate r (7.5% -> .075 in decimal), and the final P value (8500). So, we only need to solve for t.
8500 = (400)(1+.075)^t
/400 /400
21.25 = 1.075^t
logarithms are the inverse of exponents, basically, if you have an example like
y = b^x, then a logarithm inverts it, logy(baseb)=x
Makes sense if you consider a power of ten.
1000 = 10^3
if you put logbase10(1000), you'll get 3.
Anyways, though, to solve the problem make a log with a base of 1.075 in your calculator
log21.25(base 1.075) = t
also, because of rules of change of base (might want to look this up to clarify), you can write this as log(21.25)/log(1.075) = t
Thus, t is 42.26118551.
Rounded to hundredths, t=42.26
<u>Given</u>:
Given that the data are represented by the box plot.
We need to determine the range and interquartile range.
<u>Range:</u>
The range of the data is the difference between the highest and the lowest value in the given set of data.
From the box plot, the highest value is 30 and the lowest value is 15.
Thus, the range of the data is given by
Range = Highest value - Lowest value
Range = 30 - 15 = 15
Thus, the range of the data is 15.
<u>Interquartile range:</u>
The interquartile range is the difference between the ends of the box in the box plot.
Thus, the interquartile range is given by
Interquartile range = 27 - 18 = 9
Thus, the interquartile range is 9.
