Answer:
2 and 3
Step-by-step explanation:
Congruent means that they look the same. 2 and 3 are the only ones that look the same.
Answer:362
Step-by-step explanation:
You need to multiply
We have that
<span>(c-4)/(c-2)=(c-2)/(c+2) - 1/(2-c)
</span>- 1/(2-c)=-1/-(c-2)=1/(c-2)
(c-4)/(c-2)=(c-2)/(c+2)+ 1/(c-2)------- > (c-4)/(c-2)-1/(c-2)=(c-2)/(c+2)
(c-4-1)/(c-2)=(c-2)/(c+2)---------------- > (c-5)/(c-2)=(c-2)/(c+2)
(c-5)/(c-2)=(c-2)/(c+2)------------- > remember (before simplifying) for the solution that c can not be 2 or -2
(c-5)*(c+2)=(c-2)*(c-2)------------------ > c²+2c-5c-10=c²-4c+4
-3c-10=-4c+4----------------------------- > -3c+4c=4+10----------- > c=14
the solution is c=14
the domain of the function is (-∞,-2) U (-2,2) U (2,∞) or
<span>all real numbers except c=-2 and c=2</span>
'll use the binomial approach. We need to calculate the probabilities that 9, 10 or 11
<span>people have brown eyes. The probability that any one person has brown eyes is 0.8, </span>
<span>so the probability that they don't is 1 - 0.8 = 0.2. So the appropriate binomial terms are </span>
<span>(11 C 9)(0.8)^9*(0.2)^2 + (11 C 10)(0.8)^10*(0.2)^1 + (11 C 11)(0.8)^11*(0.2)^0 = </span>
<span>0.2953 + 0.2362 + 0.0859 = 0.6174, or about 61.7 %. Since this is over 50%, it </span>
<span>is more likely than not that 9 of 11 randomly chosen people have brown eyes, at </span>
<span>least in this region. </span>
<span>Note that (n C r) = n!/((n-r)!*r!). So (11 C 9) = 55, (11 C 10) = 11 and (11 C 0) = 1.</span>
