180-104=76
76x2=152
Answer is 152
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
Answer: 1 hour
Step-by-step explanation:
Let distance1 = x km
Distance2 = 144 - x
Speed S1 = 63 kph
Speed S2 = 81 kph
Time = Distance / Speed
When the jets pass each other
(144-x) / 81 = x/63
63(144-x) = 81x
9072 - 63x = 81x
x = 9072 / 144
x = 63 km
Time = 63/63 = 1 hour
The absolute value of -5 is 5.
