Answer:
The answer is 44.
Step-by-step explanation:
The distance of the spaceship in discuss as in the task content given can be evaluated as; 800miles.
<h3>What is the distance the spaceship travels in 4 minutes?</h3>
The distance travelled by the spaceship in discuss can be evaluated by means of the slope of the linear relationship as follows;
Hence it follows from ratios that by observation, the linear relationship has a slope of 200mi/min.
Consequently, we can evaluate the distance travelled after 4 minutes as;
Distance = 200 × 4 = 800mi.
Ultimately, the distance travelled per minute by the spaceship is; 800mi.
Remarks:
600 miles
520 miles
800 miles
1,080 miles
Read more on ratios;
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Complete question:
The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line
Answer:
(3.699, 4.701)
Step-by-step explanation:
Given:
Sample size, n = 45
Sample mean, x' = 4.2
Standard deviation
= 2.0
Required:
Find a 90% CI for true mean time
First find standard error using the formula:




Standard error = 0.298
Degrees of freedom, df = n - 1 = 45 - 1 = 44
To find t at 90% CI,df = 44:
Level of Significance α= 100% - 90% = 10% = 0.10

Find margin of error using the formula:
M.E = S.E * t
M.E = 0.298 * 1.6802
M.E = 0.500938 ≈ 0.5009
Margin of error = 0.5009
Thus, 90% CI = sample mean ± Margin of error
Lower limit = 4.2 - 0.5009 = 3.699
Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701
Confidence Interval = (3.699, 4.701)
The population increases by a factor 1.0167 every year so the require figure is:
103 * 1 . 0167^5 = 112 million