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katovenus [111]
2 years ago
10

I will attach a screenshot of the math problem.

Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
7 0

Complete the table for the function y = 0.1^x

The first step: plug values from the left column into the ‘x’ spot in the formula <u>y=0.1^x</u>.

* 0.1^-2 : We can eliminate the negative exponent value by using the rule a^-1 = 1/a. Keep this rule in mind for future problems. (0.1^-2 = 1/0.1 * 0.1 = 100).

* 0.1^-1 = 1/0.1 = 10

* 0.1^0 = 1 : (Remember this rule: a^0 = 1)

* 0.1^1 = 0.1

Our list of values: 100, 10, 1, 0.1

Now, we can plug these values into your table:

\left[\begin{array}{ccc}x&y\\2&10\\1&10\\0&1\\1&0.1\end{array}\right]

The points can now be graphed. I will paste a Desmos screenshot; try to see if you can find some of the indicated (x,y) values: [screenshot is attached]

I hope this helped!

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alekssr [168]

Answer:

1.25 liters of oil

Step-by-step explanation:

Volume in Beaker A = 1 L

Volume of Oil in Beaker A = 1*0.3 = 0.3 L

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If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:

V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3

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0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L

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6 0
2 years ago
If Upper X overbar equals 62​, Upper S equals 8​, and n equals 36​, and assuming that the population is normally​ distributed, c
marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)    

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=62 represent the sample mean

\mu population mean (variable of interest)

s=8 represent the sample standard deviation

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Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

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We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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