Evaluate the following limit.
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos(x)}{\cot(x)}\]
# install.packages(c("pracma","Ryacas"))
library(pracma)
library(Ryacas)
## Warning: package 'Ryacas' was built under R version 4.5.2
##
## Attaching package: 'Ryacas'
## The following object is masked from 'package:stats':
##
## integrate
## The following objects are masked from 'package:base':
##
## %*%, det, diag, diag<-, lower.tri, upper.tri
x <- ysym("x")
f <- (cos(x) / (cos(x) / sin(x)))
result <- lim(f,x,pi / 2)
result
## y: 1
Which one of the following is NOT a factor of 210?
A. 6
B. 14
C. 15
D. 16
# install.packages("tidyverse")
library(tidyverse)
## Warning: package 'lubridate' was built under R version 4.5.2
## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
## ✔ forcats 1.0.1 ✔ stringr 1.5.2
## ✔ ggplot2 4.0.0 ✔ tibble 3.3.0
## ✔ lubridate 1.9.4 ✔ tidyr 1.3.1
## ✔ purrr 1.1.0
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ purrr::cross() masks pracma::cross()
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ✖ purrr::simplify() masks Ryacas::simplify()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
q2_data <- data.frame(Choice = LETTERS[1:4],
Number = c(6,14,15,16))
correct_answer <- q2_data %>%
mutate(Factor.of.210 = 210 %% Number == 0) %>%
filter(Factor.of.210 == FALSE) %>%
pull(Choice)
cat("The correct answer is:",correct_answer,"\n")
## The correct answer is: D
A code consists of six letters chosen from the 26 letters of the alphabet. What is the chance that any two of the letters are the same?
our_letters <- LETTERS # LETTERS: built-in R constant of all capital letters
counter <- 0
N <- 1e5 # 100,000 trials
for (i in 1:N) {
letters_picks <- sample(x = our_letters,size = 6,replace = T)
if (length(unique(letters_picks)) < 6) {
counter <- counter + 1
}
}
probability <- counter / N
cat("The probability is:",probability,"\n")
## The probability is: 0.46277