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1 year ago

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Tariq wrote the equation y=3.2x to represent the line of best fit for the data shown. Is his line a good representation for the

data? Explain your reasoning.

1 answer:

Setler79 [48]1 year ago

6 0

Answer:

No

Step-by-step explanation:

The way to find the line of best fit by estimate is to have about half the points be above and below the line of best fit. In this case Tariq followed the first few points of the data but his estimate would be very off after 10 on the x axis. This would not accurately predict what the next data point could be.

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2.50x = Ethan’s total sold -6 (15/2.50=6)
2x = Chloe’s total sold
2.50(12) = 30 - 6 = 24
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They both sold 12 candy bars

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a rectangular piece of paper 71 cm long and 10m wide . a cylinder is formed by rolling the paper along its length , find volume

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Answer:

Volume=355 cm³

Step-by-step explanation:

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1 year ago

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Answer:

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Step-by-step explanation:

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The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T

Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that 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 Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue 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. In this problem, $\sigma = 0.25$

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

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

$0.1 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.1\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.1}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}$

$n = 24.01$

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

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

$0.06 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.06\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.06}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}$

$n = 66.7$

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

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

$0.05 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.05\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.05}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}$

$n = 96.04$

Rounding up, 97

8 0

1 year ago