James will end up with his original t cars and half of (t+13) cars, so will have ...
... t + (t+13/2) = (3t +13)/2 . . . . cars James has after Paul's gift
Answer:
coffee is the best answer in my mind
I see the solution in three steps.
1.) RS ⊥ ST, RS ⊥ SQ, ∠STR ≅ ∠SQR | Given
2.) RS<span>≅RS | Reflexive Property
3.) </span><span>△RST ≅ △RSQ | AAS Triangle Congruence Property</span>
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 = 17
For the alternative hypothesis,
µ < 17
This is a left tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 80,
Degrees of freedom, df = n - 1 = 80 - 1 = 79
t = (x - µ)/(s/√n)
Where
x = sample mean = 15.6
µ = population mean = 17
s = samples standard deviation = 4.5
t = (15.6 - 17)/(4.5/√80) = - 2.78
We would determine the p value using the t test calculator. It becomes
p = 0.0034
Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.
The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population mean of the length of engineering conferences.
The variable has a normal distribution.
The information given is:
n= 84
x[bar]= 3.94
δ= 1.28
The formula for the Confidence interval is:
x[bar]±
*(δ/n)
Lower bound(Lb): 3.698
Upper bound(Ub): 4.182
Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242
I hope it helps!
