kimjooin02
18.12.2019 •
Mathematics
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.]
a. log n
b. log log n
c. 100 n
d. n log n
e. n2
f. n3
g. 2n
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Ответ:
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
Ответ:
*The dot plot is shown in the attachment below
2
Step-by-step explanation:
Interquartile range is the difference between the upper median (Q3) and the lower median (Q1).
First, let's write out each value given in the data. Each dot represents a data point.
We have:
2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7
=>Find the median:
Our median is the middle value. The middle value is the 6th value = 4
==>Upper median Q3) = the middle value of the set of data we have from the median to our far right.
2, 3, 3, 4, 4, |4,| 4, 5, [5], 6, 7
Our upper median = 5
==>Lower median(Q1) = the middle value of the data set we have from our median to our far left.
2, 3, [3], 4, 4, |4,| 4, 5, 5, 6, 7
Lower median = 3
==>Interquartile range = Q3 - Q1 = 5-3 = 2