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
you have negative -12
turn this into a positive (12)
12 plus 18
= 30
Answer:
52.94
Step-by-step explanation:
61.06
-46.92
=52.94
<em>just</em><em> </em><em>subtract</em><em> </em><em>them</em><em> </em><em>like</em><em> </em><em>the</em><em> </em><em>.</em><em> </em><em>isn't</em><em> </em><em>even</em><em> </em><em>there</em><em>.</em>
Answer:
the expected value of this raffle if you buy 1 ticket = -0.65
Step-by-step explanation:
Given that :
Five thousand tickets are sold at $1 each for a charity raffle
Tickets are to be drawn at random and monetary prizes awarded as follows: 1 prize of $500, 3 prizes of $300, 5 prizes of $50, and 20 prizes of $5.
Thus; the amount and the corresponding probability can be computed as:
Amount Probability
$500 -$1 = $499 1/5000
$300 -$1 = $299 3/5000
$50 - $1 = $49 5/5000
$5 - $1 = $4 20/5000
-$1 1- 29/5000 = 4971/5000
The expected value of the raffle if 1 ticket is being bought is as follows:





Thus; the expected value of this raffle if you buy 1 ticket = -0.65