yul has lunch at a restaurant if his bill is $14.50 and yul want to leave 20% tip what is the amount he should leave for a tip
2 answers:
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Answer:
a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.
b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.
Step-by-step explanation:
Question a:
We have to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Z-table as such z has a p-value of .
That is z with a p-value of , so Z = 1.96.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.
The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.
The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.
Question b:
The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.
Answer:
<em>29 minutes more</em>
Step-by-step explanation:
Let m represent minutes
changing the statement to algebra, since the second company charges a different rate at night and weekend we have the equation below;
$19.99 + $0.35m > $29.99
Subtract 19.99 from both sides to isolate m and we have;
$19.99 -$19.99 + $0.35 > $29.99 - $19.99
= $0,35m > $10.00
Divide both side by 0.35 to obtain the value of m;
>
= m > 28.57
<em>m ⩾ 29 minutes</em>
<em>The second company's will be twenty nine minutes or more costlier than the first company</em>