Answer:
See explanations below
Step-by-step explanation:
Given the functions f(x)=2x+3 and g(x)=x^2-1
a. Find f(g(x))
f(g(x)) = f(x^2-1)
f(g(x)) = 2(x^2-1)+3
f(g(x))= 2x^2-2+3
f(g(x)) = 2x^2+1
Hence the composite function f(g(x)) is 2x^2+1
b) g(f(x)) = g(2x+3)
g(f(x) = (2x+3)^2-1
g(f(x)) = 4x^2+12x+9-1
g(f(x)) = 4x^2+12x+8
Answer:D
Step-by-step explanation:
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
I’m not *completely* sure if this would be right, but it could possibly be:
160
how i got this:
20 = 150
/(divide) 150
0.13(recurring) = 1
x1200
160 = 1200
if you need me to explain anything in detail, please feel free to message! have a nice day:)<3
Complete question :
Birth Month Frequency
January-March 67
April-June 56
July-September 30
October-December 37
Answer:
Yes, There is significant evidence to conclude that hockey players' birthdates are not uniformly distributed throughout the year.
Step-by-step explanation:
Observed value, O
Mean value, E
The test statistic :
χ² = (O - E)² / E
E = Σx / n = (67+56+30+37)/4 = 47.5
χ² = ((67-47.5)^2 /47.5) + ((56-47.5)^2 /47.5) + ((30-47.5)^2/47.5) + ((37-47.5)^2/47.5) = 18.295
Degree of freedom = (Number of categories - 1) = 4 - 1 = 3
Using the Pvalue from Chisquare calculator :
χ² (18.295 ; df = 3) = 0.00038
Since the obtained Pvalue is so small ;
P < α ; We reject H0 and conclude that there is significant evidence to suggest that hockey players' birthdates are not uniformly distributed throughout the year.
