Ok.look.the question is like that
7:2 so first of all divide 7 by 2 like that 7/2 so you get 3.5 ok than add 3.5 with 2 so the answer is 5.5
Answer:
9
Step-by-step explanation:
0.15
+3.00=3.15
3.15÷35=0.09
0.09×100=9
Answer:
a. 205320
b. 34220
c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!
Step-by-step explanation:
a) The number of ways to dustribute exams among the TA's is:
n / (n - r)!
n= number of things to choose from
r= Choosing r number
60P3= 60! / (60 - 3)!
(60)(59)(58)(57)! / (57)!
=205320
B) The number of ways to dustribute the exams among the TA's is:
n! /(n - r)! r!
60C3= 60! /(60 - 3)! 3!
= 60!/ 57! 3!
= 60 × 59 × 58 / 3 × 2 × 1
= 34220
C) The required number of ways is:
60C25 + 60C20 + 60C15
= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!
Answer:
Yes the sample can be use to make inference
Step-by-step explanation:
The inference is possible if the conditions:
p*n > 10 and q*n > 10
where p and q are the proportion probability of success and q = 1 - p
n is sample size
Then p = 12 / 30 = 0,4 q = 1 - 0,4 q = 0,6
And p*n = 0,4 * 30 = 12 12 > 10
And q*n = 0,6 * 30 = 18 18 > 10
Therefore with that sample the conditions to approximate the binomial distribution to a Normal distribution are met
