Answer:
Here we will use the relation:
Density = mass/volume.
We know that:
Mass = 1485g
Volume = 750L
Density = 1485g/750L = 1.98 g/L
We do not have the options in order to select the correct one, but knowing this density, we can suppose the possible gases.
Two "common" gases with densities similar to this one are:
Carbon Dioxide, with d = 1.977 g/L
Dinitrogen Monoxide, with d = 1.977 g/L
In both cases, if we round up we will get the same density that we calculated at the beginning, so either of these can be the correct option.
Answer:
Step-by-step explanation:
The function d(x) takes a value of x in degrees centigrade and provides the number of degrees that a container of water at that temperature x is far from the boiling point of water.
The function f(d) takes a value d in degrees centigrade and returns a value d(x) in degrees fahrenheit.
Therefore, by doing f(d(x)) we are introducing the function d(x) within the function f(d).
So the range of d(x) now is the domain of f(d(x))
This means that the function f(d(x)) shows the <em>number of degrees Fahrenheit</em> that a water container at a<em> temperature x in degrees Celsius</em> is far from the boiling point of water.
Long division: (x³ + 2) ÷ (x + 1)
<u> </u><u>x² – x + 1 </u>
x³ + 0x² + 0x + 2 | x + 1
<u>– x³ – x²</u> ⋮ ⋮
– x² + 0x ⋮
<u>+ x² + x</u><span> ⋮</span>
+ x + 2
<span> </span> <u>– x – 1</u>
+ 1
Quotient: Q(x) = x² – x – 1;
Remainder: R(x) = + 1.
I hope this helps. =)
Answer:true,true,false, true,false false
Step-by-step explanation:
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
