This would be True. For example. If your number is 24, and 'n' is 12 ( a factor of 24 )
your factors of 12 (12, 6, 4, 3, 2, 1) are also included in the factors of 24 (which are 24, 12, 6, 4, 3, 2, 1)
ASSUMING This is a straight line so we gotta the formula for a straight line which is y=mx+b, where m represents the slope and b represents the y intercept.
First, we know this line passes through (5,8) and (9,2) we can use these for finding the equations. When we know two points, we use this formula:
y-y=m(x-x)
The first y is 8 and the second one is 2
The first x is 5 and the second one is 9
Plug it in:
8-2=m(5-9)
6=m(-4)
6/-4=m <— simplify this
m= -3/2
*NOTE: another way to find m is by calculating it (y-y)/(x-x)
Now we know m, we have to find b.
All you gotta do is plug everything you know back into the equation y=mx+b
y=mx+b
y=-3/2x+b <— now plug in a point we know(x,y)
8=-3/2(5)+b
8=-15/2+b
8-(-15/2)=b
b=8+15/2
b=16/2+15/2
b=31/2 (now you can write be as a fraction or a decimal in your equation, depending on what your teacher told you to use)
*NOTE: it is best to use fractions instead of decimals as it is more accurate sometimes.
Now we know all the variables that need to be known, we just need to rewrite the formula of the equation so the teacher can see.
m=-3/2
b=31/2
We don’t need to plug in x or y since it could have different values (since a straight line has MANY co-ordinates)
SO OUR EQUATION IS=
y=(-3/2)x+31/2
Hope you understand this, feel free to ask me anything!
Answer:
a) 3.6
b) 1.897
c)0.0273
d) 0.9727
Step-by-step explanation:
Rabies has a rare occurrence and we can assume that events are independent. So, X the count of rabies cases reported in a given week is a Poisson random variable with μ=3.6.
a)
The mean of a Poisson random variable X is μ.
mean=E(X)=μ=3.6.
b)
The standard deviation of a Poisson random variable X is √μ.
standard deviation=S.D(X)=√μ=√3.6=1.897.
c)
The probability for Poisson random variable X can be calculated as
P(X=x)=(e^-μ)(μ^x)/x!
where x=0,1,2,3,...
So,
P(no case of rabies)=P(X=0)=e^-3.6(3.6^0)/0!
P(no case of rabies)=P(X=0)=0.0273.
d)
P(at least one case of rabies)=P(X≥1)=1-P(X<1)=1-P(X=0)
P(at least one case of rabies)=1-0.0273=0.9727
