Answer: 34 degrees of freedom should be used to find the p-value of the test .
Step-by-step explanation:
Degrees of Freedom relates to the maximum number of independent values, that have independence to vary in the sample.
Given : When testing the difference between two population means and the population variances are unknown and unequal, the degrees of freedom are calculated as 34.7.
But degree of freedom must be an integer , so we find the greatest integer less than equal to the calculated degree of freedom.
i.e. [df]=[34.7]= 34
Thus , 34 degrees of freedom should be used to find the p-value of the test .
The simplest form of
6
2
is
3
1
.
Steps to simplifying fractions
Find the GCD (or HCF) of numerator and denominator
GCD of 6 and 2 is 2
Divide both the numerator and denominator by the GCD
6 ÷ 2
2 ÷ 2
Reduced fraction:
3
1
Therefore, 6/2 simplified to lowest terms is 3/1.
Answer:
y = - x² - 4x - 9
Step-by-step explanation:
Given
y = - (x + 2)² - 5 ← expand (x + 2)² using FOIL
= - (x² + 4x + 4) - 5 ← distribute parenthesis by - 1
= - x² - 4x - 4 - 5 ← collect like terms
= - x² - 4x - 9 ← in standard form
The probability of getting a 5 from the first rolling is 1/6. The probability of getting an even number is 1/2 (= 1/6 (for a 2) + 1/6 (for a 4) + 1/6 (for a 6)). As the two rollings are independent, you can just multiply two probability values to come up with a final answer. That is, (1/6)·(1/2) = 1/12.
Note that I did not write this. This answer comes from Wyzant. I am only linking this to you so that you'll get the answer quickly.
Find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)
we just need the slope
take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1
the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation
solve for where y=x-1 and y=x-x² intersect
set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero
x-1=0
x=1
we got this point already
x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2
it intersects at (-1,-2)
