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HACTEHA [7]
3 years ago
6

Find three solutions of the equation y = 7x – 5.

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
5 0
 y=7x-5 This is the same as saying y-7x = -5 Well easy. You just make up numbers that make this work. Let x be 1 y-7=-5 y=-5+7 y=2 So one ordered pair is (1,2) Let x=2 y-14=-5 y=9 So another ordered pair is (2,9) Let x=3 y-21=-5 y=-5+21 y=16 So another ordered pair is (3,16) 
hope it helps
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VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a p-value of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

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3 years ago
Joe makes $13.35 an hour. If Joe works 40 hours in one week, how much would he earn for the week?
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Answer:

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Step-by-step explanation:

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