Answer:

Step-by-step explanation:
Two lines are given to us which are perpendicular to each other and we need to find out the value of a . The given equations are ,
Step 1 : <u>Conver</u><u>t</u><u> </u><u>the </u><u>equations</u><u> in</u><u> </u><u>slope</u><u> intercept</u><u> form</u><u> </u><u>of</u><u> the</u><u> line</u><u> </u><u>.</u>
and ,
Step 2: <u>Find </u><u>the</u><u> </u><u>slope</u><u> of</u><u> the</u><u> </u><u>lines </u><u>:</u><u>-</u>
Now we know that the product of slope of two perpendicular lines is -1. Therefore , from Slope Intercept Form of the line we can say that the slope of first line is ,
And the slope of the second line is ,
Step 3: <u>Multiply</u><u> </u><u>the </u><u>slopes </u><u>:</u><u>-</u><u> </u>
Multiply ,
Multiply both sides by a ,
Divide both sides by -1 ,
<u>Hence </u><u>the</u><u> </u><u>value</u><u> of</u><u> a</u><u> </u><u>is </u><u>9</u><u> </u><u>.</u>
God bless you ma’am/ sir.!!!!-8hsiwhwhjwiwis&/&):8292jwiaiia21161
You can compute both the mean and second moment directly using the density function; in this case, it's
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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)
m<1=152
m<2=28
m<3=152
m<4=28
m<5=152
m<6=28
m<7=152
not completly sure but good luck
Answer:
im not sure but i think its 48
Step-by-step explanation:
if you divide the two you get 96. Divide that by the average rate number (2) and get 48.