Answer:
second option
Step-by-step explanation:
Given
+ 6x² + 5 = 0
let u = x², then
u² + 6u + 5 = 0 ← in standard form
(u + 1)(u + 5) = 0 ← in factored form
Equate each factor to zero and solve for u
u + 1 = 0 ⇒ u = - 1
u + 5 = 0 ⇒ u = - 5
Change u back into terms of x, that is
x² = - 1 ( take the square root of both sides )
x = ±
= ± i ( noting that
= i ), and
x² = - 5 ( take the square root of both sides )
x = ±
= ±
= ±
×
= ± i
Solutions are x = ± i and x = ± i
Answer: A) .1587
Step-by-step explanation:
Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
i.e.
and 
Let x denotes the amount of soda in any can.
Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.
Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =
![P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587](https://tex.z-dn.net/?f=P%28x%3E12.50%29%3D1-P%28x%5Cleq12.50%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B12.50-12.30%7D%7B0.20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1%29%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.8413%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%5C%5C%5C%3D0.1587)
Hence, the required probability= A) 0.1587
You first must find the 18% of the fish population.
You multiply 500 by .18 . You are supposed to get 90.
90 is the 18% fish increase.
Then you just add 500 + 90 and you will get 590 fish that are now in the school.
Answer:
120
Step-by-step explanation:
If 6 = 5%, then 6 = 1/20 of the total grade. 6×20= 120