My answer is 8 5/9
First, you have to multiply the denominators to get your least common denominator.
Second, you have to multiply the numerators and the denominators together to get two different fractions with the same denominator. You should get 9 3/9 and 1 6/9.
Your answer should be 8 5/9
Answer:
<u>2/5 < 5/8 < 6/7 < 1 </u>
<u>OR</u>
<u>1 > 6/7 > 5/8 > 2/5</u>
Step-by-step explanation:
It is required to compare Two-fifths, Six-sevenths, Five-eighths, and 1
Two-fifths = 2/5
Six-sevenths = 6/7
Five-eighths = 5/8
So, the given numbers are: 2/5, 6/7, 5/8, and 1
We need to make the numbers in order from the least to the greatest or from the greatest to the least
The easy method is convert the rational numbers to decimal numbers
So,
2/5 = 0.4
6/7 ≈ 0.857
5/8 = 0.625
1 = 1
So, the numbers form the least to the greatest are:
0.4 , 0.625 , 0.857 , 1
So,
2/5 , 5/8 , 6/7 , 1
The inequality correctly compares the numbers are:
<u>2/5 < 5/8 < 6/7 < 1</u>
Or can be written from the greatest to the least as:
<u>1 > 6/7 > 5/8 > 2/5 </u>
Answer:
Its B
Step-by-step explanation:
I remember doing this :)
Answer:
a) P=0.558
b) P=0.021
Step-by-step explanation:
We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.
The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:


b) The approximate probability that at least 9 carry the gene is:


