AT&T has a family plan of four lines for $125 and additional lines are $15 each per month. If Jason wants to spend at most $
200 per month on cell phone expenses, how many additional lines can he afford? Which inequality represent the given situation. please help for my final
1 answer:
Given:
Cost of four lines = $125
Cost of each additional line = $15
Jason wants to spend at most $200 per month on cell phone expenses.
To find:
The inequality for the given situation.
Solution:
Let be the number of additional line.
Cost of one additional line = $15
Cost of additional line =
Total cost = Fixed cost + Addition cost
=
It is given that Jason wants to spend at most $200 per month on cell phone expenses. It means the total cost must be less than or equal to 200.
Therefore, the correct option is C.
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Answer:
[0.7210,0.8189] = [72.10%, 81.89%]
Step-by-step explanation:
The sample size is
n = 200
the proportion is
p = 154/200 = 0.77
<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>
<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference. </em>
The approximation would be to a Normal curve with this parameters:
<em>Mean </em>
p = 0.77
<em>Standard deviation </em>
The 90% confidence interval for the proportion would be then
where is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval
is 10%=0.1
We can either use a table, a calculator or a spreadsheet to get this value.
In Excel or OpenOffice Calc we use the function
<em>NORMSINV(0.95) and we get a value of 1.645 </em>
The 90% confidence interval for the proportion is then
This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%
If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?
Yes, I do.