Gardeners on the west coast of the united states are investigating the difference in survival rates of two flowering plants in drought climates. plant a has a survival rate of 0.64 and plant b has a survival rate of 0.42. the standard error of the difference in proportions is 0.084. what is the margin of error for a 99% confidence interval?

use critical value z = 2.576.

moe = round all calculations to three decimal places.

The 99% confidence interval would be given (0.004;0.436).  

Step-by-step explanation:

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 A  

represent the estimated proportion A

represent the real population proportion B  

represent the estimated proportion B

represent the critical value for the margin of error  

The population proportion have the following distribution  

The confidence interval for the difference of two proportions 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.  

The standard error is given by:

And replacing into the confidence interval formula we got:  

And the 99% confidence interval would be given (0.004;0.436).  

answer: idk

step-by-step explanation:

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer