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iragen [17]
2 years ago
12

The function y=kx is given in the table. Find the coefficient k and fill in the table.

Mathematics
2 answers:
Sloan [31]2 years ago
8 0

In the 1st table of values,

take x = 3 and y = 12

since,

y = kx

k = 12÷3

k = 4.

thus,

x = -2, y = -8

x = -1/3, y = -4/3

x = 0, y = 0

y = 16 , x = 3

y = 1/8, x = 1/32

All the best!

rodikova [14]2 years ago
6 0

Answer:

yo im doing the test rn lol wish me luck :D

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Data given and notation

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\hat p=0.45 estimated proportion of chips that fail in the first 1000 hours of their use

\mu_0 =0.48 is the value that we want to test

\alpha=0.05 represent the significance level

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Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

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z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion  is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

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It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.  

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