Answer:
D. d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
C. d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
B. d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Step-by-step explanation:
Given two points (x1, y1) and (x2,y2) we can find the distance using
d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
The order of the terms inside the square doesn't matter
d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
When we are squaring are term, we can take the absolute value before we square and it does not change the value
d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Okay there is going to be a 90°, 45°, and a 45° atleast
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ±
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71

[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
Answer:
0.
Step-by-step explanation:
Using the laws of logarithms:
log81/8 + 2log2/3 - 3log 3/2 + log 3/4
= log 81/8 + log (2/3)^2 - log (3/2)^3 + log 3/4
= log 81/8 + log 4/9 - log 27/8 + log 3/4
= log 81/8 + log 4/9 - (log 27/8 - log 3/4)
= log (81/8 * 4/9) - log (27/8 * 4/3)
= log 9/2 - log 9/2
= 0.