Answer:
We accept H₀. We do not have argument to keep the claim that the mean breaking strength has increased
Step-by-step explanation:
Normal Distribution
Population Mean μ₀ = 1850 pounds
Standard Deviation σ = 90 pounds
Type of test
Null Hypothesis H₀ ⇒ μ = μ₀
Alternative Hypothesis Hₐ ⇒ μ > μ₀
A one tail test (right)
n = 21 as n < 30 we use t-student table
degree of fredom 20
t = 2.845
Sample mean μ = 1893
Then, we compute t statistics
t(s) = [ 1893 - 1850 ] / 90/ √n
t(s) = 43 * 4,583 / 90
t(s) = 197,069 / 90
t(s) = 2,190
And we compare t and t(s)
t(s) = 2.190
t = 2.845
Then t(s) < t
We are in the acceptance zone, we accept H₀
You can do 20/2=10 This helps because it is just doing the reverse so you know if you get it both ways that it is correct. Medal please?
A. $0.02 per golf tee
To check this answer;
0.02 * 150 golf tees = $3
Let me know if this helps :)
Answer:
Harry is 2, Jim is 6
Step-by-step explanation:
The first statement tells us they are 4 years part. Then you need to find two numbers that are 4 apart and that has one that is half the other, like 4 and 8. Finally, subtract two because this will happen in two years and you need their ages now.
Sorry if this was confusing.
So point-slope form is (y-y1) = m(x-x1). So first find m. m = (y2-y1) / (x2-x1) So for your problem m = (-4-2) / (1--8) = -6/9 = -2/3. Then plug the slope and a point into the point-slope equation so you can either get: y-2 = (-2/3)*(x+8) or y+4 = (-2/3)*(x-1). Both are valid.
Hope this helps!
