The equation has one extraneous solution which is n ≈ 2.38450287.
Given that,
The equation;

We have to find,
How many extraneous solutions does the equation?
According to the question,
An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.
To solve the equation cross multiplication process is applied following all the steps given below.

The roots (zeros) are the  x  values where the graph intersects the x-axis. To find the roots (zeros), replace  y 
with  0  and solve for  x. The graph of the equation is attached.
n  ≈  2.38450287
Hence, The equation has one extraneous solution which is n   ≈  2.38450287
For more information refer to the link.
brainly.com/question/15070282
 
        
             
        
        
        
Answer:
The z-score for the 34-week gestation period baby is 0.61
Step-by-step explanation:
The formula for calculating a z-score is is z = (x-μ)/σ, 
where x is the raw score,
 μ is the population mean
σ is the population standard deviation.
We are told in the question that:
Babies born after a gestation period of 32-35 weeks have a mean weight of 2600 grams and a standard deviation of 660 grams. Also, we are supposing a 34-week gestation period baby weighs 3000grams 
 
The z-score for the 34-week gestation period baby is calculated as:
z = (x-μ)/σ
x = 3000, μ = 2600 σ = 660
z = 3000 - 2600/660
= 400/660
=0.6060606061
Approximately, ≈ 0.61
 
        
             
        
        
        
Step-by-step explanation:
 just multiple them and you'll get the results and the results you got add them together
 
        
             
        
        
        
Answer:
The car must have a speed of 25 kilometres per hour to stop after moving 7 metres. 
Step-by-step explanation:
Let be  , where
, where  is the stopping distance measured in metres and
 is the stopping distance measured in metres and  is the speed measured in kilometres per hour. The second-order polynomial is drawn with the help of a graphing tool and whose outcome is presented below as attachment.
 is the speed measured in kilometres per hour. The second-order polynomial is drawn with the help of a graphing tool and whose outcome is presented below as attachment.
The procedure to find the speed related to the given stopping distance is described below:
1) Construct the graph of  .
.
2) Add the function  .
.
3) The point of intersection between both curves contains the speed related to given stopping distance.
In consequence, the car must have a speed of 25 kilometres per hour to stop after moving 7 metres.