Answer: 2:16 - 3:24 - 4:32 - 5:40 - 6:48 - 7:56 - 8:64 // just keep adding each side to its own side
Step-by-step explanation:
Well I don't know the answer choices so
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. 30x + 20
2. -10x + 60
Step-by-step explanation:
Answer:
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Step-by-step explanation:
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we are given with an isosceles triangle with two congruent sides equal to 70 inches and an angle of 36 degrees in between the sides. We are asked for the area of the triangle. The formula is A = 0.5 * ab sin theta where a and b are the sides. The area is equal to 1440. 07 in2.
