If you call 
the mass of the ant and 
the load, we have the equation
In fact, the mass of the ant is one tenth of the load, which is exactly what this equation states.
Since we are given the load, we simply need to plug its value in the equation to deduce the mass of the ant:
Vertical Asimtote: (x-3) this will be the denominator
Horizontal Asymtote: (x-1)^2 (x+3) this will be the numerator
y= a (x-1)^2 (x+3) / (x-3)
Vertical: (x-1/2)
Horizontal: (x-3)
y= a (x-1/2)/(x-3)
I hope this helped?
Notice that we can simplify both numerator and denominator of our rational function. In the numerator we have a quadratic expression of the form
. To to simplify it, we are going to find tow numbers that add to 2 and multiply to -8; those numbers are 4 and -2.
In the denominator we have a difference of squares:
Now we can rewrite our function:
From the simplified form of our rational function we can infer that its graph has two vertical asymptotes at
and
We can conclude that the graphic of our rational function is:
Answer:
$18,007,50
Step-by-step explanation:
First, you have to calculate the 85% of the base price that the dealer pays for the car:
base price: $18,750
$18,750*85%= $15,937.5
Second, you have to calculate the 75% of the installed options price that the dealer pays:
installed options price= $2,380
$2380*75%= $1,785
Third, you have to add the 85% of the base price plus the 75% of the installed options that the dealer has to pay and you also have to add the destination charge of $285:
$15,937.5+$1,785+$285= $18,007.5
According to this, the dealer has to pay $18,007.5 for the car with a base price of $18,750 and installed options price $2380 including a destination charge of $285.
Answer:
It would change to 0.04802
Step-by-step explanation:
from this question we have that n became 400
40% of 400
= 160
p* = 160/400
= 0.4
1 - p* =
= 1 - 0.4
= 0.6
at confidence level,
1 - 0.95
= 0.05
alpha/2 = 0.025
z= 1.96
<u>margin of error. E</u>
= 1.96 x √[(0.4 x 0.6)/400]
= 1.96 x 0.0245
= 0.04802
M.E = 0.04802
