Answer:
I would rather use their multiples.
Step-by-step explanation:
With 13 and 14, neither number has many factors (and few prime factors, for that matter) so factoring would be pretty much useless. If you wanted to use the prime factors of 14 (7 and 2) multiplying either of them by 13 would not give you the LCM. Actually the LCM is just 13*14 which is 182.
Good luck!
Answer:
a) 5/21
b) 4/21
c) 4/21
d) 8/21
Step-by-step explanation:
total number of coins: 21
a) number if dollars: 5
therefore fraction is 5/21
b) number of quarters: 4
therefore fraction is 4/21
c) number of dimes: 4
therefore fraction is 4/21
d) number of nickels: 8
therefore fraction is 8/21
Answer:
90 in²
Step-by-step explanation:
The figure's area is that of four right triangles, each with legs of 6 in and 7.5 in. The area of each triangle is half the product of the leg lengths, so is ...
triangle area = (1/2)(6 in)(7.5 in)
Then the area of 4 of those triangles is ...
figure area = 4 · triangle area = 2(6 in)(7.5 in) = 90 in²
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
& 
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
= 
= 
= 0.4
Part c)
Assume the events are not independent
Given Data
P
= 0.4
=
= 0.4
= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P
= 1 - 
= 1 - ![[P({A}')+P({B}') - P({A}'\cap {B}')]](https://tex.z-dn.net/?f=%5BP%28%7BA%7D%27%29%2BP%28%7BB%7D%27%29%20-%20P%28%7BA%7D%27%5Ccap%20%7BB%7D%27%29%5D)
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56
Answer:
504
Step-by-step explanation:
