If 5 consecutive integers is 205,
then a + b + c + d + e = 205
but also, each integer is separated by a difference of 1
⇒ a + (a + 1) + (a + 2) + (a + 3) + (a + 4) = 205
⇒ 5a + 10 = 205
⇒ 5a = 195
⇒ a = 39
∴ third term = 39 + 2
= 41
Answer:
0.2364
Step-by-step explanation:
We will take
Lyme = L
HGE = H
P(L) = 16% = 0.16
P(H) = 10% = 0.10
P(L ∩ H) = 0.10 x p(L U H)
Using the addition theorem
P(L U H) = p(L) + P(H) - P(L ∩ H)
P(L U H) = 0.16 + 0.10 - 0.10 * p(L u H)
P(L U H) = 0.26 - 0.10p(L u H)
We collect like terms
P(L U H) + 0.10P(L U H) = 0.26
This can be rewritten as:
P(L U H)[1 +0.1] = 0.26
Then we have,
1.1p(L U H) = 0.26
We divide through by 1.1
P(L U H) = 0.26/1.1
= 0.2364
Therefore
P(L ∩ H) = 0.10 x 0.2364
The probability of tick also carrying lyme disease
P(L|H) = p(L ∩ H)/P(H)
= 0.1x0.2364/0.1
= 0.2364
Answer:
b should be correct
Step-by-step explanation:
no need to cash app me its been a while since ive done math so i dont even know if this is right
6.5 + 7.6
8.4 + 5.7
ur welcome
