Answer:
1)0.123( bar on 3)
Let X = 0.123 ( bar on 3)
Then X = 0.1233 --------(1)
Multiply equation (1) by 100 we get,
100X = 12.33 --------(2)
Again multiply equation (2) by 10 we get,
1000X = 123.33 -------(3)
Subtract equation (2) from equation (3) we get,
1000X = 123.33
100X = 12.33
____________
900X = 111
X = 111/900
Hence,
0.123 ( bar on 3) is in the form of 111/900 which is in the form of P/Q.
Assuming a normal distribution we find the standardized z scores for a:-
z1 = (80 - 180) / 25 = = -100/25 = -4
z2 = (280-180) / 25 = 4
Required P( -4 < z < 4) from the tables is >99.9%
b
z1 = 130-180 / 25 = -2
z2 = 230-180 / 25 = 2
from tables probability is 2* 0.4773 = 95.46 %
Answer: no yes
Step-by-step explanation:
The answer is C have a good day.-
