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sleet_krkn [62]
3 years ago
7

A bottle contains 24.5 oz of water. A leak in the bottle causes it to lose 0.3 oz of water every minute. How many minutes does i

t take for the bottle to lose 33% of the water?
Mathematics
2 answers:
Bogdan [553]3 years ago
8 0

Answer:

There is 1.5 water left in the bottle

Step-by-step explanation:

Alona [7]3 years ago
7 0

Answer: 54.7 Minutes

Step-by-step explanation:

33% × 24.5 =

(33 ÷ 100) × 24.5 =

(33 × 24.5) ÷ 100 =

808.5 ÷ 100 =

8.085 ≈

8.09;

so then 8.09 - 24.5 = 16.41

16.41 divided by 0.3 = 54.7

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

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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)\\\\

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

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