Answer: 13 and 31 sorry for it being rog the first time
Step-by-step explanation: BRAINLIEST PLZ
Answer: X=15
Step-by-step explanation: Isolate the variable by dividing each side by factors that don't contain the variable.
Hope this helps!! :)
Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
Answer:
0.2322 or 23.22 %
Step-by-step explanation:
We have to solve and find the area out of these limits
μ + 0,3 = 210 + 0,3 ⇒ 210,3 and
μ - 0,3 = 210 - 0,3 ⇒ 209.7
z(l) = ( x - 210 ) / (2.8/√84) ⇒ z(l) = - (0.3 * 9,17)/ 2.8
z (l) = - 1.195
We need to interpole from z table
1.19 ⇒ 0.1170
1.20 ⇒ 0.1151
Δ ⇒ 0.01 ⇒ 0.0019
And between our point 1,195 and 1,19 the difference is 0.005
then 0.01 ⇒ 0.0019
0.005 ⇒ ?? (x)
we find x = 0.00095
to get the area for poin z (l) - 1.195 up to final left tail is from z table
0,1170 - 0.00095 = 0.1161
And by symmetry to the right is the same
So 0.1161 * 2 = 0.2322
We find the area out of the above indicated limits the area we were looking for. This is the probability of finding shafts over and below the population mean and 0.3 inches
Step-by-step explanation:
Answer:
0
Step-by-step explanation:
Any value multiplied by zero equals zero , thus
0 × | - 6 | = 0
