1 = 10
2 = 12
3 = 14
4 = 16
5 = 18
6 = 20
Δweight
------------- =
Δ cost
2-1
------- =
12-10
1
—
2
Call me 205 579 7799 if u need help because this stuff is easy
X < 6
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A. 9 3/4 hours
b. i’m not too sure :)
Answer:
p=![\frac{181}{301}](https://tex.z-dn.net/?f=%5Cfrac%7B181%7D%7B301%7D)
Step-by-step explanation:
30100 have dogs, 18100 have cats.
The question simply asks for a union scenario thus the law of AND & OR is applicable.
Therefore:-Those who have both cats and dogs partially have cats.
![P(X=both \ cat \ and \ dog)=\frac{18100}{30100}\\p=\frac{181}{301}](https://tex.z-dn.net/?f=P%28X%3Dboth%20%5C%20cat%20%5C%20and%20%5C%20dog%29%3D%5Cfrac%7B18100%7D%7B30100%7D%5C%5Cp%3D%5Cfrac%7B181%7D%7B301%7D)