Answer:
4.5341 < 6.9 < 6.906 < 6.96
Step-by-step explanation:
4.5341 is closer to 0 than 6.96 is.
Answer:
the answer for the question is 10+x
y-3
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: 42.21 km
Step-by-step explanation:
We can solve this using trigonometry, since we have the following data:
is the the angle of elevation
is the horizontal distance between the plane and the radar station
is the hypotenuse of the right triangle formed between the radar station and the airplane
Now, the trigonometric function that will be used is <u>cosine</u>:
because 
is the adjacent side of the right triangle
Finding :
<span>x−<span>2/7</span></span>=<span>5/<span>7
so this means x = 1</span></span>
