Autores:
Colaborador:
Se necesita calcular lo valores numéricos aproximados de los siguientes problemas.
pro2<-(0.3*0.15)/((0.3*0.15)+(0.2*0.8)+(0.5*0.12))
pro2
## [1] 0.1698113
pro3<-(5**6/factorial(6))*exp(-5)
pro3
## [1] 0.1462228
pro4<- (factorial(20)/(factorial(7)*factorial(13))*(2/5)**7 * (3/5)**13)
pro4
## [1] 0.1658823
Un estudiante debe realizar las siguientes sumas.
1+2+3+…+10001+2+3+…+1000
Sol.: 500500500500
a<-sum(1:1000)
a
## [1] 500500
c<-c(a,a)
c
## [1] 500500 500500
suma0<-0
n<-0
repeat{
suma0<-suma0+n
n<-n+1
if(n>=1001)break
}
suma0
## [1] 500500
t<-c(suma0,suma0)
t
## [1] 500500 500500
suma1<-0
n<-0
while(n<=1000){
suma1<-suma1+n
n<-n+1
}
suma1
## [1] 500500
totalsuma<-c(suma1,suma1)
totalsuma
## [1] 500500 500500
1+2+4+8+16+…+10241+2+4+8+16+1024
Sol.: 20472047
sum<-0
n<-1
while(n<=1024){
sum<-sum+n
n<-n*2
}
sum
## [1] 2047
totalsum<-c(sum,sum)
totalsum
## [1] 2047 2047
Colocar el nombre de los 20 alumnos de un curso en un vector y realizar lo que se le solicite.
alumnos<-c("Alison","Javi","Fred","Sofía","Cami",
"Paco","Juan","Tami","Sami","Pame",
"Abii","Daniela","Alex","Dami","Gene",
"Luis","Isac","Anna","Luay","Rosa")
alumnos
## [1] "Alison" "Javi" "Fred" "Sofía" "Cami" "Paco" "Juan"
## [8] "Tami" "Sami" "Pame" "Abii" "Daniela" "Alex" "Dami"
## [15] "Gene" "Luis" "Isac" "Anna" "Luay" "Rosa"
length(alumnos)
## [1] 20
a<-substring(alumnos,1,1)
which(a=="A")
## [1] 1 11 13 18
Después de crear el vector alumnos
con los nombres de
los 20 estudiantes, ahora necesitamos crear un vector llamado
notas
el cual guardará las notas de cada uno de los
estudiantes en el mismo orden que se encuentran los nombres.
alumnos<-c("Alison","Javi","Fred","Sofía","Cami",
"Paco","Juan","Tami","Sami","Pame",
"Abii","Daniela","Alex","Dami","Gene",
"Luis","Isac","Anna","Luay","Rosa")
notas<-c(9,8,7.7,8,6,7,9,7,7.7,6,10,3,6,5,7.8,7,8,10,8,9)
notas
## [1] 9.0 8.0 7.7 8.0 6.0 7.0 9.0 7.0 7.7 6.0 10.0 3.0 6.0 5.0 7.8
## [16] 7.0 8.0 10.0 8.0 9.0
length(notas)
## [1] 20
suma<-sum(notas)
suma
## [1] 149.2
mean(notas)
## [1] 7.46
which(notas>7)
## [1] 1 2 3 4 7 9 11 15 17 18 19 20
sort(notas)
## [1] 3.0 5.0 6.0 6.0 6.0 7.0 7.0 7.0 7.7 7.7 7.8 8.0 8.0 8.0 8.0
## [16] 9.0 9.0 9.0 10.0 10.0
sort(notas,decreasing=T)
## [1] 10.0 10.0 9.0 9.0 9.0 8.0 8.0 8.0 8.0 7.8 7.7 7.7 7.0 7.0 7.0
## [16] 6.0 6.0 6.0 5.0 3.0
max(notas)
## [1] 10
which(notas>=10)
## [1] 11 18
Con los vectores alumnos
y notas
ya creados
ahora realizaremos lo siguiente:
s<-head(notas,10)
sum(s)
## [1] 75.4
length(s)
## [1] 10
sum(s)
## [1] 75.4
nota2 = notas
notas
## [1] 9.0 8.0 7.7 8.0 6.0 7.0 9.0 7.0 7.7 6.0 10.0 3.0 6.0 5.0 7.8
## [16] 7.0 8.0 10.0 8.0 9.0
nota2
## [1] 9.0 8.0 7.7 8.0 6.0 7.0 9.0 7.0 7.7 6.0 10.0 3.0 6.0 5.0 7.8
## [16] 7.0 8.0 10.0 8.0 9.0
nota2[ notas < 7 ] = 'Reprobado'
nota2[ notas >= 7 ] = 'Aprobado'
nota2
## [1] "Aprobado" "Aprobado" "Aprobado" "Aprobado" "Reprobado" "Aprobado"
## [7] "Aprobado" "Aprobado" "Aprobado" "Reprobado" "Aprobado" "Reprobado"
## [13] "Reprobado" "Reprobado" "Aprobado" "Aprobado" "Aprobado" "Aprobado"
## [19] "Aprobado" "Aprobado"
which(nota2=="Aprobado")
## [1] 1 2 3 4 6 7 8 9 11 15 16 17 18 19 20
aprob<-c(which(nota2=="Aprobado"))
aprob
## [1] 1 2 3 4 6 7 8 9 11 15 16 17 18 19 20
length(aprob)
## [1] 15
porc<-(length(aprob)/20)*100
cat(porc,"%")
## 75 %
max(s)
## [1] 9
min(s)
## [1] 6
(q<-cbind(s,alumnos))
## s alumnos
## [1,] "9" "Alison"
## [2,] "8" "Javi"
## [3,] "7.7" "Fred"
## [4,] "8" "Sofía"
## [5,] "6" "Cami"
## [6,] "7" "Paco"
## [7,] "9" "Juan"
## [8,] "7" "Tami"
## [9,] "7.7" "Sami"
## [10,] "6" "Pame"
## [11,] "9" "Abii"
## [12,] "8" "Daniela"
## [13,] "7.7" "Alex"
## [14,] "8" "Dami"
## [15,] "6" "Gene"
## [16,] "7" "Luis"
## [17,] "9" "Isac"
## [18,] "7" "Anna"
## [19,] "7.7" "Luay"
## [20,] "6" "Rosa"
which.max(s)
## [1] 1
which.min(s)
## [1] 5
mean(aprob)
## [1] 10.4