Write an equation of a line that passes through the point (2, 3), parallel to y=12x−3 1) y=1/2x+3 2) y=1/2x+2 3) y=2x+3 4) y=2x+
1 answer:
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Answer:
2) D: x = [0, 24]
3) R: y = [0, 384]
4) see graph
<u>Step-by-step explanation:</u>
Eric's regular wage is $12 per hour for all hours less than 9 hours.
The minimum number of hours Eric can work each day is 0.
f(x) = 12x for 0 ≤ x < 9
Eric's overtime wage is $18 per hour for 9 hours and greater.
The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).
f(x) = 18(x - 8) + 12(8)
= 18x - 144 + 96
= 18x - 48 for 9 ≤ x ≤ 24
The daily wage where x represents the number of hours worked can be displayed in function format as follows:
2) Domain represents the x-values (number of hours Eric can work).
The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.
D: 0 ≤ x ≤ 24 → D: x = [0, 24]
3) Range represents the y-values (wage Eric will earn).
Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.
The minimum hours he can work in one day is 0.
f(x) = 12x
f(0) = 12(0)
= 0
The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>
f(x) = 18x - 48
f(24) = 18(24) - 48
= 432 - 48
= 384
D: 0 ≤ y ≤ 384 → D: x = [0, 384]
4) see graph.
Notice that there is an open dot at x = 9 for f(x) = 12x
and a closed dot at x = 9 for f(x) = 18x - 48
Answer:
The estimation for the proportion for this case is
And we know that the 99% confidence interval is given by:
So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:
If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:
d. 21% ± 4.8%
Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
The population proportion have the following distribution
The confidence interval would be given by this formula
For the 95% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
For the 99% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
The estimation for the proportion for this case is
And we know that the 99% confidence interval is given by:
So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:
If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:
d. 21% ± 4.8%
Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)