Count the number of positive integers less than 100 that do not contain any perfect square factors greater than 1.
Possible perfect squares are the squares of integers 2-9.
In fact, only squares of primes need be considered, since for example, 6^2=36 actually contains factors 2^2 and 3^2.
Tabulate the number (in [ ])of integers containing factors of
2^2=4: 4,8,12,16,...96 [24]
3^2=9: 9,18,....99 [11]
5^2=25: 25,50,75 [3]
7^2=49: 49,98 [2]
So the total number of integers from 1 to 99
N=24+11+3+2=40
=>
Number of positive square-free integers below 100 = 99-40 = 59
Answer:
And in the figure attached we see the limits with the percentages associated.
Step-by-step explanation:
For this case we know that the random variable of interest is the scores on a test given to all juniors in a school district follows a normal distribution with the following parameters:
For this case we know from the empirical rule that within one deviation from the mean we have approximately 68.2% of the data, within 2 deviations from the mean we have 95% and within 3 deviation 99.7%
We can find the limits and we got:
And in the figure attached we see the limits with the percentages associated.
Answer: adults or pre teens
Step-by-step explanation:
Answer:
x=37°
y=6
Step-by-step explanation:
(ASSUMING BD is a perpendicular bisector of AC, otherwise answers maybe be wrong)
Consider the triangle ABC:
AB is the same length as BC (marked), meaning ABC is an isosceles triangle.
Since ABC is isosceles, ∠BAC=∠BCA=53
All angles in a triangle add up to 180:
180=∠ABC + ∠BAC + ∠BCA
180= ∠ABC + 53 +53
∠ABC = 180 - 106
∠ABC = 74
Assuming BD bisects ∠ABC perfectly, x is half of ∠ABC.
x=74/2=37
If BD bisects AC perfectly, AD=DC=y=6
Step-by-step explanation:
Below is an attachment containing the solution
