<h2>Answer:</h2><h3>=60</h3><h2>Step-by-step explanation:</h2><h3>=Solution:</h3><h3> (8+2) × (8-2)</h3><h3> =10×6</h3><h3> =60#</h3>
3(x - 25) < $100. We're not req'd to solve this.
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
Consider the function

, which has derivative

.
The linear approximation of

for some value

within a neighborhood of

is given by

Let

. Then

can be estimated to be

![\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B63.97%7D%5Capprox4-%5Cdfrac%7B0.03%7D%7B48%7D%3D3.999375)
Since

for

, it follows that

must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function

. This means the estimated value is an overestimation.
Indeed, the actual value is closer to the number 3.999374902...
To find the interquartile range, you will list the data that is presented in the stem and leaf plot.
Find the median of the data (30.5)
Find the median of the lower half and the median of the upper half.
Subtract these two values.
The data are <u>20</u>, 25, 30, 30, 31, 40, 41, <u>49</u>.
27.5 40.5
40.5-27.5 = 13
The interquartile range is 13.