Answer:
the probability of an adult getting a NEG result and truly having tuberculosis is 0.0127
Step-by-step explanation:
S = the adult really has tuberculosis.
S' = complement of S = the adult does not has tuberculosis.
POS = the test gives a positive result
P(S)= 0.05
P(POS | S)=0.746
P(NEG | S')= 0.7653
this is an intersection because the "and" word
P(NEG ∩ S) = P(NEG| S)*P(S)=(1-P(POS | S))*P(S)=(1-0.746)*0.05=0.0127
Area of parallelogram<span>=</span><span>base x </span><span>height
Given :
Base = 12 in.
Height = 3 in.
Hence,
Area = 12 x 3
= 36 in. square</span>
Answer:
"Absolute value" is the non-negative value of a number, disregarding whether or not it has a sign.
Step-by-step explanation:
"Absolute value" is the non-negative value of a number, disregarding whether or not it has a sign. It can be thought of as how far it is away from the number of 0, whether it's to the left or right. Absolute value of a number is written as |x|, where x = any number. Let's say x was -1. |-1| would be 1 because it's 1 away from 0. If x was just 1, then the number would stay the same.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
Substitute
, so that
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%2B%5Cdfrac1x%5Cleft%28%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D%5Cright%29%3D%5Cdfrac1%7Bx%5E2%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%29)
Then the ODE becomes
which has the characteristic equation
with roots at
. This means the characteristic solution for
is
and in terms of
, this is
From the given initial conditions, we find
so the particular solution to the IVP is
Answer:
678
Step-by-step explanation:
Okay I'll try to help you. :)
