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:
1.False
2.False
3.True
4.True
5.not sure, it could be true or false because adjacent angles can add up to 180 or even 360 as long as they have the same vertex
Hello,
I have a graphing calculator so this will be easy for me to answer and explain how to figure these out. Here we go!
1. Well on a Ti - 84 plus calculator it says graph then you put in thee equations for each and C and D are already incorrect cause they go straight up and B is incorrect because it is a straight line on the top going straight. In this case your looking for a line that goes through the dots. And the answer would be A) Y = 1/2x + 7.
2. B) Then umber of accidents were reduced by 0.67 per month for every additional driver in the program.
3. Key: Y means yes N means no. Plot A had 1 disease present and 2 not present. Plot B has 3 diseases present and 1 disease not present.
4. Category A had 36 groups in group 1. and Category B had 12 groups in group 1. Easy!
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Hope I Helped!
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