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ankoles [38]
2 years ago
12

Kelli swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 6 minut

es. If the river flows at 5 km/hr, how fast can Kelli swim in still water?
Choose the most logical value for the variable to represent.
Let x = .

3m = 15

9 + 15m = 12m – 1

3m = 18

9 + 6 – 15 = 15m – 12m
Mathematics
1 answer:
never [62]2 years ago
5 0

By solving a system of equations, we conclude that Kelli's speed on still water is 6.1km/h

<h3></h3><h3>How fast can Kelli swim in still water?</h3>

Let's say that the speed of Kelli in still water is S. And we know that the speed of the river is 5km/h.

When she swims upstream, her velocity is:

(S - 5km/h)

When she swims downstream, her velocity is:

(S + 5km/h)

With the given information, we can write for a distance D:

(S - 5km/h)*1h = D

(S + 5km/h)*6min = D

First, let's rewrite 6 minutes into hours.

1 hour = 60min

6 min = 6/(60) hours = 0.1 hours

Then the system of equations that we need to solve is:

(S - 5km/h)*1h = D

(S + 5km/h)*0.1h = D

D is the same distance in both cases, then we can write:

(S - 5km/h)*1h = (S + 5km/h)*0.1h

Now we can solve this for S.

S*1h - 5km = S*0.1h + 0.5km

S*1h - S*0.1h = 5km + 0.5km

S*(0.9h) = 5.5km

S = 5.5km/0.9h = 6.1km/h

Kelli's speed on still water is 6.1km/h

If you want to learn more about systems of equations:

brainly.com/question/13729904

#SPJ1

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Step-by-step explanation:

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Answer:

a) The null hypothesis is

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b) Check Explanation.

c) The $3000 is obviously the cost of a type I error.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and is always about the absence of significant difference between two proportions being compared. It usually maintains that random chance is responsible for the outcome or results of any experimental study/hypothesis testing. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared. It usually confirms the the theory being tested by the experimental setup. It usually maintains that other than random chance, there are significant factors affecting the outcome or results of the experimental study/hypothesis testing. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to verify that the mean diameter produced by the machine is really off-target.

Note that, that target is 3 inches.

Hence, the null hypothesis will be that that there is no difference between the mean diameter of the 100 cylinders sampled and the target of 3 inches.

And the alternative hypothesis will confirm the concerns of the technical staff that there is a significant difference between the mean diameter of the 100 cylinders sampled and the target diameter of 3 inches. That is, the mean diameter is off-target.

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The null hypothesis is

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b) A type I error involves rejecting the null hypothesis and accepting the alternative hypothesis when in reality, the null hypothesis is true. It involves saying there is significant evidence to show that the mean diameter of the sampled cylinders is indeed different from the targeted 3 in. (that is, the mean diameter of sampled cylinders is off-target), so they try to fix the issue when in reality, there is actually no significant difference between the mean diameter of sampled cylinders and the targeted 3 inches.

While a type II error involves failing to reject the null hypothesis when in reality it should have been rejected.

It entails not rejecting the null hypothesis and making conclusions based on the null hypothesis, when in reality, the alternative hypothesis should have been accepted together with its conclusion.

In this one the firm would conclude that they do not need to change anything as there is no significant difference between the mean diameter of the sampled cylinders and the targeted 3 inches, when in reality, there is a significant difference between the mean diameter of the sampled cylinders and the targeted 3 inches diameter.

c) The $3000 is obviously the cost of a type I error.

This is because the type I error is the one that involves tryin to fix the problem that did not even exist in reality.

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