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vesna_86 [32]
2 years ago
7

Hello, could someone please help me find three pairs for this question

Mathematics
1 answer:
amid [387]2 years ago
7 0

Answer:

x - y = -3        x + y = 9

x      y            x     y

-2 , 1              -2 , 11

-1 , 2              -1 , 10

0 , 3               0 , 9

1 , 4                1 , 8

2 , 5                2 , 7

Hope this helps.

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Oh lordy I hope this is correct

Step-by-step explanation:

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3 years ago
Hope researches the impact of one variable on another. She graphs the data from her research and sees that the data forms a shap
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She can determine that while there is an association between the variables, there is no correlation, and she cannot determine causation.

Step-by-step explanation:

Since there is an arch shape to her graph, we know that as one variable changes, the other changes in the same manner. This means there is an association between the variables.

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3 years ago
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"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

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