Answer:
it is 1/2
Step-by-step explanation:
The product of a lines slope M and the slope K of a perpendicular is -1:
M*K=-1
This implies
K=-1/M
since
M * -1/M=-1
In our example we are given a line of slope M=-2 therefore the slope K of a line to perpendicular to this line is:
M*K=-1
-2*K=-1
K=-1/-2=1/2
Ten times (10 x) the sum (+) of half a number (
![\frac{n}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%7D%7B2%7D%20)
) and 6 (6) is 8 (=8)
10(<span>
![\frac{n}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%7D%7B2%7D%20)
+ 6) = 8 Use the Distributive Property
</span>
![\frac{10n}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B10n%7D%7B2%7D%20)
+ 60 = 8 Simplify the fraction
<span> 5n + 60 = 8 Subtract 60 from both sides
5n = -52 Divide both sides by 5
n = -10</span>
Answer:
hello :
note : secy = 1 / cosy
tany = siny /cosy
sin y+cos y+tan y sin y = sin y+cos y+( siny /cosy)sin y
= ( siny cosy + cos²y + sin²y) / cosy
= (siny cosy +1) / cosy......because : cos²y+sin²y =1
= (siny cosy) / cosy + 1 / cosy
= (siny/cosy) × cosy + 1/cosy
sin y+cos y+tan y sin y = tany cosy + secy
Answer:
a) P(X≥143)=0
b) This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).
Step-by-step explanation:
If we use the normal approximation to the binomial distribution we have the following parameters (mean and standard deviation):
![\mu=np=0.68*206=140.1\\\\ \sigma=\sqrt{np(1-p)}=\sqrt{206*0.68*0.32}=\sqrt{44.8256}=6.7](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D0.68%2A206%3D140.1%5C%5C%5C%5C%20%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B206%2A0.68%2A0.32%7D%3D%5Csqrt%7B44.8256%7D%3D6.7)
Then, we can calculate the probability of X being equal or more than 143 using the z-score:
![z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{143-140.1}{6.7/\sqrt{206}}=\dfrac{2.9}{0.4668}=6.2124\\\\\\P(x\geq143)=P(z>6.2124)=0](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B143-140.1%7D%7B6.7%2F%5Csqrt%7B206%7D%7D%3D%5Cdfrac%7B2.9%7D%7B0.4668%7D%3D6.2124%5C%5C%5C%5C%5C%5CP%28x%5Cgeq143%29%3DP%28z%3E6.2124%29%3D0)
This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).