3 blocks east and 2 blocks north should be the correct answer :)
Answer:
the probability that five randomly selected students will have a mean score that is greater than the mean achieved by the students = 0.0096
Step-by-step explanation:
From the five randomly selected students ; 160, 175, 163, 149, 153
mean average of the students = 160+175+163+149+153/5
= mean = x-bar = 800/5
mean x-bar = 160
from probability distribution, P(x-bar > 160) = P[ x-bar - miu / SD > 160 -150.8 /3.94]
P( Z>2.34) = from normal Z-distribution table
= 0.0096419
= 0.0096
hence the probability that five randomly selected students will have a mean score that is greater than the mean achieved by the students = 0.0096
where SD = standard deviation = 3.94 and Miu = 150.8
Answer:
84
Step-by-step explanation:
Answer:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
