Consider the number below. How many times less is the value of

The scale on a map shows as 1:5000000. What would be the length on the map if the length of a road in

SVEN [57.7K]

Answer:

70 cm

Step-by-step explanation:

$\frac{1}{5 * 10^{6} } = \frac{d}{350} \\\\d = \frac{350}{5 * 10^{6} } \\\\d = 7 * 10^{-5} km = 7 * 10^{-5} * 10^{6} cm = 70 cm$

6 0

3 years ago

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

motikmotik

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$