Find (dy/dx) when x=2at,y=at^2
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A function assigns the values. The correct option is C, x=-6.
<h3>What is a Function?</h3>A function assigns the value of each element of one set to the other specific element of another set.
A.) When the value of x is -12,
f(-12) = 2/3(-12) + 7 = -1
g(-12) = 2(-12) + 15 = -9
B.) When the value of x is -9,
f(-9) = 2/3(-9) + 7 = 1
g(-9) = 2(-9) + 15 = -3
C.) When the value of x is -6,
f(-6) = 2/3(-6) + 7 = 3
g(-6) = 2(-6) + 15 = 3
D.) When the value of x is -3,
f(-9) = 2/3(-3) + 7 = 5
g(-9) = 2(-2) + 15 = 11
Hence, the correct option is C, x=-6.
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Answer:
a)
And the 99% confidence interval would be given (0.630;0.730).
b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)
c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population
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".
represent the real population proportion of interest
represent the estimated proportion
n=575 is the sample size required (variable of interest)
represent the critical value for the margin of error
The population proportion have the following distribution
Part a
The confidence interval would be given by this formula
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.
And replacing into the confidence interval formula we got:
And the 99% confidence interval would be given (0.630;0.730).
Part b
We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)
Part c
For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population