Give each month a number:
 January = 0
February = 1
 March = 2
April = 3
Now set X to 3 in the equation ad solve for t.
t = -30cos(x/6) +60
t = -30cos(3/6) +60
t = 33.672 degrees. ( Round as necessary)
You would need to change the +60 to a new starting point based on what the rise in temperature is due to global warming.
 
        
             
        
        
        
The maximum profit is the y-coordinate of the vertex of the parabola represented by the equation.

The maximum value is 6400, but the profit is given in hundreds of dollars, so multiply the value by 100.
The maximum profit the company can make is $640,000.
 
        
        
        
If it is a square, subtract (90+ whatever the other angle is) from 360. If it is a triangle than add 90 plus whatever other number the other angle is then subtract it from 180. A triangle is 180 degrees, and a square is 360 degrees.
        
             
        
        
        
Answer:
y=2x+3
Step-by-step explanation:
the normal charge is 3 + 2 x number of days so this equation is the correct one.
 
        
             
        
        
        
You can compute both the mean and second moment directly using the density function; in this case, it's

Then the mean (first moment) is
![E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B80%7D%5Cint_%7B670%7D%5E%7B750%7Dx%5C%2C%5Cmathrm%20dx%3D710)
and the second moment is
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B80%7D%5Cint_%7B670%7D%5E%7B750%7Dx%5E2%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7B1%2C513%2C900%7D3)
The second moment is useful in finding the variance, which is given by
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2%3D%5Cdfrac%7B1%2C513%2C900%7D3-710%5E2%3D%5Cdfrac%7B1600%7D3)
You get the standard deviation by taking the square root of the variance, and so
![\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%3D%5Csqrt%7B%5Cdfrac%7B1600%7D3%7D%5Capprox23.09)