Hello there!
By multiplying them altogether, this would the result in (-0.054).
We first do the first two problems.
0.04*-0.9.
From this solution, we multiply this by 1.5, and from this we get -0.054
The number 61 is larger because it a bigger number than 13 and 13 is smaller because it more smaller number than 61
Answer:
μ−2σ = 1,089.26
μ+2σ = 1,097.62
Step-by-step explanation:
The standard deviation of a sample of size 'n' and proportion 'p' is:
![\sigma=\sqrt{\frac{p*(1-p)}{n} }](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%2A%281-p%29%7D%7Bn%7D%20%7D)
If n=1139 and p =0.96, the standard deviation is:
![\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%2A%281-p%29%7D%7Bn%7D%7D%5C%5C%5Csigma%20%3D%200.001836)
The minimum and maximum usual values are:
![\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n](https://tex.z-dn.net/?f=%5Cmu-2%5Csigma%20%3D%20%28p-2%5Csigma%29%2An%5C%5C%5Cmu%2B2%5Csigma%20%3D%20%28p%2B2%5Csigma%29%2An)
![\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62](https://tex.z-dn.net/?f=%5Cmu-2%5Csigma%20%3D%20%280.96-2%2A0.001836%29%2A1139%5C%5C%5Cmu-2%5Csigma%20%3D%201%2C089.26%5C%5C%5Cmu%2B2%5Csigma%20%3D%20%280.96%2B2%2A0.001836%29%2A1139%5C%5C%5Cmu%2B2%5Csigma%20%3D%201%2C097.62)
Make a table with the angle theta as independent variable and the radius r as dependent variable:
theta radius = 4+2cos theta radius
------- -----------------------------------------
0 4+2 6
pi/6 4+2cos pi/6 = 4+2(sqrt(3)/2
Perhaps you have already plotted this using webassign (but remember that you have not shared an illustration here). (Please don't type "webassign plot" repeatedly, as it accomplishes nothing.)
Generally, when one wishes to find the area of a region defined by polar functions (as is the case here), one first determines suitable limits of integration from the finished curve and checks them through actual integration.
Which formula should you use to find the area: Look up "areas in polar coordinates," as I did. The formula is as follows:
Enclosed area = Integral from alpha to beta of (1/2)r^2 d(theta). Note that the initial radius here is 6 (since r = 4 plus 2 cos theta is 4+2 when theta = 0).
Assuming Lindsey takes 5 hours to mow a lawn, she earns from twelve lawns
12*70=$840.
If the above assumption is not correct, @funnyork2005 needs to clarify the question.