Question 1181618: Modern oil tankers weigh more than a 1/2 million tons and have lengths up to one - fourth mile. Such massive ships require a distance of 5.0 km about (3.0 mi) and a time of 20 minutes to come to a stop from top speed of 30 km/h.what is the magnitude of such ship's average acceleration in m/s squared in coming to a stop?
B. What is the magnitude of the ship's average velocity in m/s
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's a complete hypothesis test for the manufacturer's claim:
**a) Solution:**
We will use a two-sample z-test for the difference between means since the sample sizes are large (n > 30) and the population standard deviations are known.
**b) Parameter under study:**
The parameter under study is the difference between the average tensile strength of yarn A (μA) and the average tensile strength of yarn B (μB). We are interested in whether μA - μB < 12.
**c) Statement of Hypotheses:**
* Null hypothesis (H₀): μA - μB ≥ 12 (The difference is greater than or equal to 12 kg)
* Alternative hypothesis (H₁ or Ha): μA - μB < 12 (The manufacturer's claim: The difference is less than 12 kg)
**d) Significance level:**
α = 0.005
**e) Test statistic:**
The test statistic for a two-sample z-test is:
z = [(xA - xB) - (μA - μB)₀] / sqrt(σA²/nA + σB²/nB)
Where:
* xA and xB are the sample means
* (μA - μB)₀ is the hypothesized difference under H₀ (12)
* σA and σB are the population standard deviations
* nA and nB are the sample sizes
z = [(86.7 - 77.8) - 12] / sqrt((6.28²/50) + (5.61²/45))
z = [-3.1] / sqrt(0.788 + 0.700)
z = -3.1 / sqrt(1.488)
z ≈ -3.1 / 1.22
z ≈ -2.54
**f) Rejection region:**
Since this is a one-tailed test (less than), we look for the critical z-value that corresponds to α = 0.005 in the left tail of the standard normal distribution. Using a z-table or calculator, we find that the critical z-value is approximately -2.576.
The rejection region is z < -2.576.
**g) Decision making:**
Our calculated test statistic (z ≈ -2.54) falls *within* the rejection region (z < -2.576) .
Therefore, we *reject* the null hypothesis.
**h) Conclusion:**
There is sufficient evidence at the 0.005 significance level to support the manufacturer's claim that the average tensile strength of yarn A is less than the average tensile strength of yarn B by less than 12 kilograms.
Answer by ikleyn(52775)  (Show Source):
You can put this solution on YOUR website! .
Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it looks like that this guy, @CPhill, even does not look at
and does not read what he posts, so his posts are often irrelevant or incorrect.
All @CPhill' solutions are copy-paste of Google AI solutions.
But there is one essential difference.
Every time, Google AI acknowledges at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions,
but @CPhill never prints this acknowledge and never say that his solutions are copy-paste that of Google AI.
So, in my view, doing this way, this guy, @CPhill, simply makes dishonest business at this forum.
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