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2 years ago

System A

Mathematics

1 answer:

Sonbull [250]2 years ago

4 0

For the given systems we have:

1) Swap the right sides of both equations in system A.

2) No, the systems are not equivalent systems.

<h3>How can we get System B from System A?</h3>

The two systems are:

-10x - 4y = 0

-2x + 7y = 8

-10x - 4y = 8

-2x + 7y = 0

Notice that the only different thing between the systems is that the right sides are inverted. So, if we invert the right side of A, we will get system B.

2) Are the systems equivalent?

No, the systems are not equivalent because to go from system A to system B we only modified the right side of system A, the systems would be equivalent only if modifying both sides of A gave system B, but this is not the case.

If you want to learn more about systems of equations:

brainly.com/question/13729904

#SPJ1

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Charlotte walks home from school every day at a rate of

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Answer:

Noah takes 20 minutes to get to his house, that is, 10 minutes less than Charlotte.

Step-by-step explanation:

Since Charlotte walks home from school every day at a rate of 4 miles per hour and it takes her 30-minutes to reach home, while her brother, Noah, runs home from the same school every day at a rate of 6 miles per hour, to determine how many fewer minutes does it take Noah than Charlotte to reach home from school each day the following calculation must be performed:

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3 years ago

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3 years ago

The Toylot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A) under a norma

kotegsom [21]

Answer:

Test statistic = 3.1587

Step-by-step explanation:

We are given that the Toy-lot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A).

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So, Null Hypothesis, $H_0$ : $\mu$ = 0.8 { claim of 0.8 A is not low}

Alternate Hypothesis, $H_1$ : $\mu$ > 0.8 { claim of 0.8 A is too low}

Now, the test statistics used here will be;

T.S. = $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, X bar = sample mean = 1.20 A

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n = sample size = 11 motors

So, Test statistics = $\frac{1.20 - 0.8}{\frac{0.42}{\sqrt{11} } }$ ~ $t_1_0$

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At 1% level of significance, t table gives a critical value of 2.764 at 10 degree of freedom. Since our test statistics is higher than the critical value so we have sufficient evidence to reject null hypothesis .

Therefore, we conclude that Toy-lot claim of 0.8 A is too low.

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