Answer:A)35choices
B)475,020choices
C)132choices
D)0.000278
E)0.000166
Step-by-step explanation:
Given 7poultry items represented by 7p
10 red meat items=10R
12 vegetarian items=12V
@
A)there are 7combination 3ways of selecting 3out of 7 poultry items
From the formula nCr=n!/(n-r)!r!
7C3=7!/(7-3)!3!=7!/4!×3!=35choices
b)total items available=7p+10R+12V(denotation explained above)=29items
Choosing 6items implies
29C6=29!/(29-6)!×6!
29!/23!×6!=475,020 ways
C)7C2+10C2+12C2
7!/5!2!+10!/8!2!+12!/10!2!
21+45+66=132 choices
D) probability (choosing 2items from each class)=
7C2+10C2+12C2/29C6
Where 29C6 represent total probable.
21+45+66/29!/23!6!
=132/475020=0.000278choices
E)for meats to be in the majority on a 6choices selection impliee that vegetarian items will be
0,1 or2i.e combination of vegetarian and meat will be(0,6)or(1,5)or(2,4) .
Prob(0veg)+prob(1veg item) or prob(2veg items)
12C0+12C1+12C2/29C6
=1+12+66/475,020
=0.000166
Answer: b
explanation: only choice B equals -9 like the given equation.
Answer:
First, second, third, and fourth are functions
Step-by-step explanation:
Each input in the first, second, third, and fourth only have one output. In the fifth one 3.3 and 6.6 are seen twice as the x-value, so this is incorrect.
(answered this on your other question, but I will put it here too!)
Cada entrada en la primera, segunda, tercera y cuarta tiene solo una salida. En el quinto, 3.3 y 6.6 se ven dos veces como el valor de x, por lo que esto es incorrecto.
(respondí esto en tu otra pregunta, ¡pero también lo pondré aquí!)