Answer:
a) ![\cos(\theta) = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b) ![\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) ![\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d)![\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Step-by-step explanation:
We will use the following trigonometric identities
.
Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that
which implies that ![x=-\sqrt[]{49-16} = -\sqrt[]{33}](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B%5D%7B49-16%7D%20%3D%20-%5Csqrt%5B%5D%7B33%7D)
. Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then
![\cos(\theta) = \frac{-\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b)Recall that ![\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}](https://tex.z-dn.net/?f=%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2C%20%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D)
, then using the identity from above, we have that
![\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Csin%28%5Ctheta%29%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%2B%5Ccos%28%5Calpha%29%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B4%7D%7B7%7D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) Recall that 
. Then,
![\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Ccos%28%5Ctheta%29%5Ccos%28%5Cpi%29%2B%5Csin%28%5Ctheta%29%5Csin%28%5Cpi%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Ccdot%28-1%29%20%2B%200%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d) Recall that 
and ![\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%28%5Ctheta%29%7D%3D%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D)
. Then
![\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Ctan%28%5Ctheta%29%2B%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%7B1-%5Ctan%28%5Ctheta%29%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Answer is 1.25 and converted into a fraction is 5/4 converted into a mixed fraction is 1 1/4
Answer:
Well, these simulation are based on the statistics (lognormal-distributed PE, χ²-distributed s²). If you believe that only the ‘gold-standard’ of subject-simulations are valid, we can misuse the function sampleN.scABEL.sdsims() – only for the 3- and 4-period full replicates and the partial replicate:
# define a reg_const where all scaling conditions are ‘switched off’
abe <- reg_const("USER", r_const = NA, CVswitch = Inf,
CVcap = Inf, pe_constr = FALSE)
CV <- 0.4
2x2x4 0.05 0.4 0.4 0.95 0.8 1.25 34 0.819161 0.8
Since the sample sizes obtained by all simulations match the exact method, we can be confident that it is correct. As usual with a higher number of simulations power gets closer to the exact value.
Step-by-step explanation:
Answer: 3,266.53
Sorry if it’s wrong
Answer:
1 5/11 ≈ 1.455
Step-by-step explanation:
The total number of vehicles owned is
(4×0 + 8×1 + 6×2 + 4×3) = 32
The total number of families is ...
4 + 8 + 6 + 4 = 22
Then the mean number of vehicles per family is ...
32/22 = 16/11 ≈ 1.455
