Answer:
There is a 100% probability that the selected tick is also a carrier of Lyme disease.
Step-by-step explanation:
The problem states that:
Of the ticks that carry at least one of these diseases in fact carry both of them.
So, if a tick carries one of these diseases, there is a 100% probability that is carries another.
What is the probability that the selected tick is also a carrier of Lyme disease?
There is a 100% probability that the selected tick is also a carrier of Lyme disease.
Answer:
x=4/3
Step-by-step explanation:
2+4=2(x-1) original
6=2(x-1) combine like terms
6=2x-2 multiply x-1 by 2
8=2x add 2 to both sides
4=x divide both sides by 2
X=1, Y=-2. Heres your answer. It should be C
Step-by-step explanation:
d(v)=1.1(45)+0.6(45)²
=1.1(45)+0.6(2025)
=49.5+1215
=1264.5
This is all a bit complicated so try and stick with me on this one!
This one is the first problem in the first picture! a + 2 / a^2 + a - 1 / a + 1/ -a^2 - 2a - 1
= a^4 + a^3 - a^2 - a / -a^5 - 3a^4 - 3a^3 - a^2
= -a^3 - a^2 + a + 1 / a^4 + 3a^3 + 3a^2 + a
= -a^3 - a^2 + a + 1/ a^4 + 3a^3 + 3a^2 + a
= (-a - 1) (a + 1) (a - 1) / a(a+ 1) (a + 1) (a + 1)
= -a + 1 / a^2 + a
The second problem in the first picture! 3x / y + 3x - y^2 / 3xy - 9x^2 + y^2 + 9x^2 / y^2 - 9x^2
= -81x^4y + 18x^2y^3 - y^5 / 243x^5 - 54x^3y^2 + 3xy^4
This one is for the last picture! 4y^2 + 4y + 1 / 4y - 8y^2 - 4y^2 + 1 / 4y + y
= 16y^3 + 24y^2 / -32y^3 + 16u^2
= 16y + 24 / -32y + 16
= 2y + 3 / -4y + 2
I hope this was helpful!!!
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