Cross-multiply so that 8(3x + 1) = 14(2x)
Distribute the 8 and 14 into the parentheses:
24x + 8 = 28x
Subtract 24x from both sides:
8 = 4x
Divide both sides by 4:
x = 2
#1 15cm^2. ====> Good Luck! <====
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)
Answer:
Test statistic:
z = (0.67 - 0.63) / 0.033 = 1.21
p - Value = P(z > 1.21) = 0.106
My conclusion: Fail to reject null hypothesis
No, there is no evidence of a difference
It might be d but i’m not sure