Answer:
?
Step-by-step explanation:
Since both numbers are in the radical just multiply and get sq rt
25 times 4 =100
√100=10 answer
Answer:
D
Step-by-step explanation:
In order to solve this we need to plug in the answer choices and make sure they work for BOTH answers!
Let us start with option A. (6,3) 3 = 4(6) + 6 = 3 = 30 This is automaticly wrong because 3 doesnt equal 30. There is no need to try it on the second equation since the first one was wrong.
Next answer choice, B (3,6) 6 = 2(3) = 6=6. The second equation works perfect for this. Now we have to make sure that it works for the first one as well. 6 = 4(3) + 6 = 6 = 18 which is wrong.
Next, C. (1,2) 2 = 2(1) = 2=2 Right now lets check the other equation.
2 = 4(1) + 6 = 2=10. Wrong
Last, D. (-3, -6) -6 = 2(-3) = -6 = -6 Right now lets check the other equation.
-6 = 4(-3) + 6 = -6 = -6. CORRECT. Answer choice D works for both equations.
Sorry for the wait of response.
Answer:
Step-by-step explanation:
The first 6 rows of the eruptions data :
eruptions waiting
1 3.600 79
2 1.800 54
3 3.333 74
4 2.283 62
5 4.533 85
6 2.883 55
R code :
1. You can directly access the "Faithful" data in R without importing the data. The dataset faithful is present in the R or you can load the datasets. or use install the datasets.load. package
If you have the data in a text file, make sure all the columns and rows are separated by commas
Step 1: open notepad
Step 2: enter data with no spaces but only commas
Step 3: save the file as ‘faithful.txt’ on your Desktop
# Get R help
?read.table
# Import the data
rain<-read.table("C:/Users/YOUR-NAME/Desktop/faithful.txt", header = TRUE,
sep = ",")
Check the data
data("faithful") #Loading Faithful data
head(faithful, 6) #Reading first 6 rows of the data
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.
