Step-by-step explanation:
Distance = positive value
a + a = 2a
T= sqrt(5/wc)
remove ( )
T= 5sqrt/wc
multiply by wc
Twc=5sqrt
divide by Tc
W= 5sqrt/cT
hope this helps
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 25235
For the alternative hypothesis,
µ > 25235
This is a right tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 100,
Degrees of freedom, df = n - 1 = 100 - 1 = 99
t = (x - µ)/(s/√n)
Where
x = sample mean = 27524
µ = population mean = 25235
s = samples standard deviation = 6000
t = (27524 - 25235)/(6000/√100) = 3.815
We would determine the p value using the t test calculator. It becomes
p = 0.000119
Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.
Answer:
supplementary angles
Step-by-step explanation:
Answer:
See deduction below
Step-by-step explanation:
I will use the known inference rules (modus ponens, etc)
From d) and b),
~r
q → r
Therefore ~q (by Modus Tollens)
From a), and our previous conclusion:
p ∨ q
~q
Therefore p (by disjunctive sillogism)
Until know, we have concluded p and ~q. By e)
~q → u ∧ s
~q
Therefore u∧s. (Modus Ponens)
From p, u∧s, and c)
u∧s
s (simplification)
p (previous conclusion)
p∧s (adjuntion)
p∧s→t (Modus Ponens)
Therefore t, as we wanted to conclude.
