Answer:
1) [tex]f(x)=8(x-\frac{1}{4})^2 +\frac{21}{2}[/tex]
2) The new mean is 67.33
3) The vertex form is [tex]\boxed{f(x)=15(x+2)^2-79}[/tex]
The vertex for the function is: [tex]\boxed{(-2,-79)}[/tex]
4) The median is 21
[tex]Q_3=22[/tex]
Step-by-step explanation:
1) The given function is :
[tex]f(x)=8x^2-4x+11[/tex]
We want to write this function in vertex form:
First we factor 8 from the first two terms to get:
[tex]f(x)=8(x^2-\frac{1}{2} x)+11[/tex]
Next add and subtract the square of half the coefficient of the linear term
[tex]f(x)=8(x^2-\frac{1}{2} x+\frac{1}{16} )+11-8*\frac{1}{16}[/tex]
We factor the perfect square trinomial to get:
[tex]f(x)=8(x-\frac{1}{4})^2 +\frac{21}{2}[/tex]
2) We have that, the mean of 50 numbers is 68.
This implies that:
[tex]\sum x=50*68=3400[/tex]
Two numbers 82 and 86 were deleted because they are outliers.
The new sum is [tex]3400-82-86=3232[/tex]
Now the new mean is [tex]\bar X=\frac{3232}{50-2}=\frac{3232}{48}=67.3333[/tex]
The new mean is therefore 67.33 to the nearest hundreth
3)
The given function is :
[tex]f(x)=15x^2+60x-19[/tex]
We factor 15 to get:
[tex]f(x)=15(x^2+4x)-19[/tex]
Add and subtract the square of half the coefficient of x.
[tex]f(x)=15(x^2+4x+4)-19-4*15[/tex]
[tex]f(x)=15(x^2+2)^2-79[/tex]
The vertex form is [tex]\boxed{f(x)=15(x+2)^2-79}[/tex]
The vertex for the function is: [tex]\boxed{(-2,-79)}[/tex]
4) The given data is in ascending order is 14,16,19,20,21,21,22,23.
The median is 19 because it is the middle number of the array.
The first quartile is 17.5
The third quartile is 22
The interquartile range is 4.5
The box plot is shown in the attachment.
The true statements are:
The median is 21
[tex]Q_3=22[/tex]
