Troy and Lazetta are solving the following word problem. Rosalie's cat weights 9.8 pounds and her dog weighs 25.4 pounds. What i
1 answer:
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Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.
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Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote
I would write it as
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For problem 3, we first need to convert the runner's speed from mph to feet per second.
Since the runner needs to travel 90-12 = 78 ft, this means
So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.
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The question is now: how much of a lead does the runner need in order to beat the throw?
Well the runner needs to get to second base in under 0.96 seconds.
Let's calculate the distance based on that, and based on the speed we calculated earlier above.
This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet
This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.
As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.
Answer:
3.90% probability that the third failure will occur on the tenth component tested
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not fail. Components perform independently. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Assume that a component passes a test is 0.85
So they fail with probability of
What is the probability that the third failure will occur on the tenth component tested
First 9 components: Two failures, that is, P(X = 2) when n = 9.
10th component: Failure with probability 0.15.
So
In which
So
3.90% probability that the third failure will occur on the tenth component tested