<em>Greetings from Brasil...</em>
The average for a set of 9 elements will be
(A + B + C + D + E + F + G + H + I) ÷ 9 = 20
Let's make (A + B + C + D + E + F + G + H + I) like S
<em>(I chose S to remember a sum)</em>
Let us think.....
S ÷ 9 = 20
S = 20 × 9
S = 180
So, (A + B + C + D + E + F + G + H + I) = 180
According to the statement, we will include a number (element J) in the sum to obtain a mean of (20 - 4), that is:
<h3>(A + B + C + D + E + F + G + H + I +
J) ÷ 10 = (20 - 4)</h3>
as seen above, (A + B + C + D + E + F + G + H + I) = 180, then
(180 + J) ÷ 10 = 16
(180 + J) = 160
J = 160 - 180
<h2>J = - 20</h2><h2 />
So, including the number - 20 <em>(minus 20)</em> in the original mean we will obtain a new mean whose result will be 16
Answer:
90%
Step-by-step explanation:
90 + 10 is 100 total
90/100 is just 90%
Answer:
Answer is 6x-13
Step-by-step explanation:
Given
XZ=8x-18 and RZ=2x+5
XR=XZ-RZ
8x-18-2x+5
8x-2x-18+5
6x-13Ans
Answer:
f(2n)-f(n)=log2
b.lg(lg2+lgn)-lglgn
c. f(2n)/f(n)=2
d.2nlg2+nlgn
e.f(2n)/(n)=4
f.f(2n)/f(n)=8
g. f(2n)/f(n)=2
Step-by-step explanation:
What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]
from a
f(n)=logn
f(2n)=lg(2n)
f(2n)-f(n)=log2n-logn
lo(2*n)=lg2+lgn-lgn
f(2n)-f(n)=lg2+lgn-lgn
f(2n)-f(n)=log2
2.f(n)=lglgn
F(2n)=lglg2n
f(2n)-f(n)=lglg2n-lglgn
lg2n=lg2+lgn
lg(lg2+lgn)-lglgn
3.f(n)=100n
f(2n)=100(2n)
f(2n)/f(n)=200n/100n
f(2n)/f(n)=2
the time will double
4.f(n)=nlgn
f(2n)=2nlg2n
f(2n)-f(n)=2nlg2n-nlgn
f(2n)-f(n)=2n(lg2+lgn)-nlgn
2nLg2+2nlgn-nlgn
2nlg2+nlgn
5.we shall look for the ratio
f(n)=n^2
f(2n)=2n^2
f(2n)/(n)=2n^2/n^2
f(2n)/(n)=4n^2/n^2
f(2n)/(n)=4
the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare
6.n^3
f(n)=n^3
f(2n)=(2n)^3
f(2n)/f(n)=(2n)^3/n^3
f(2n)/f(n)=8
the ratio will be times 8 the initial
7.2n
f(n)=2n
f(2n)=2(2n)
f(2n)/f(n)=2(2n)/2n
f(2n)/f(n)=2
The first step is to write each factor in expanding notation.
This is:
- 124 = 100 + 20 + 4
- 2 = 2
Now muliply 2 times each term of the terms 100, 20 and 4
=> 2 * 100 = 200
2 * 20 = 40
2 * 4 = 8
Then,
(100 + 20 + 4 )
x 2
-----------------------
8
40
200
------------------------
248
