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Andre45 [30]
1 year ago
13

(Scientific Notation in the Real World MC)

Mathematics
1 answer:
Brrunno [24]1 year ago
5 0

The average number of viewers over the two week is D. 4.85 x 10^5.

<h3>How to calculate the value?</h3>

It should be noted that the scientific notation simply has to do with writing of numbers that are either too large or too small. This will be illustrated thus:

In this case, the TV show had 5.6 x 10^5 viewers in the first week and 4.1 x 10^5 viewers in the second week.

Therefore, the average number of viewers over the two weeks will be:

= (5.6 × 10^5) + (4.1 × 10^5) / 2

= 4.85 × 10^5

Note that we had to divide by 2 since it's average.

Learn more about notation on:

brainly.com/question/5756316

#SPJ1

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Q1. Butternut is a ski resort in Massachusetts. One of their triple chair lifts unloads 576 skiers per hour at the top of the sl
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As an estimation we are told 5 miles is 8 km. <br> Convert 92 km to miles.
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In a study to compare two different corrosion inhibitors, specimens of stainless steel were immersed for four hours in a solutio
djverab [1.8K]

Answer:

(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862

(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138

And the 95% confidence interval for the difference in the means is given by: 10.862 \leq \mu_A -\mu_B \leq 33.138

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_A = 242 sample mean for inhibitor A

s_A = 20 sample standard deviation for inhibitor A

n_A = 47 sample size for A

\bar X_B = 220 sample mean for inhibitor B

s_B = 31 sample standard deviation for inhibitor B

n_B = 42 sample size for A

Solution to the problem

For this case the confidence interval for the difference of means is given by:

(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}

The degrees of freedom are given by:

df = n_A +n_B -2 = 47+42-2= 87

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,87)".And we see that t_{\alpha/2}=1.988

And replacing we got:

(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862

(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138

And the 95% confidence interval for the difference in the means is given by: 10.862 \leq \mu_A -\mu_B \leq 33.138

4 0
3 years ago
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