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tekilochka [14]
3 years ago
11

A bag holds 6 green, 9 red and 4 blue marbles a marble is selected at random what is the probability of not selecting a red marb

le?
Mathematics
2 answers:
Butoxors [25]3 years ago
5 0

Answer:

The probability of not selecting a red marble, P(R)'  = 10/19

Step-by-step explanation:

The probability of an event occurring is equal to the number of ways the required event occurs divided by the number of possible outcomes

The number of ways of an event not occurring is 1 less the probability of the event occurring

The number of green marbles in the bag = 6

The number of red marbles in the bag = 9

The number of blue marbles in the bag = 4

The sum of the marbles in the bag, n = 6 + 9 + 4 = 19

The number of ways of selecting a red marble = 9 ways

The probability of selecting a red marble = 9 ways/19 marbles = 9/19

The probability of not selecting a red marble = 1 - 9/19 = 10/19 = 0.\overline{526315789473684210}

The probability of not selecting a red marble P(R)' = 10/19

Ket [755]3 years ago
4 0

Answer:

The probability of NOT selecting a red marble is 10/19 (or 52.6%)

Step-by-step explanation:

6 green+ 9 red+ 4 blue = 19 marbles in total. So 19 would be our denominator. The total of marbles that IS NOT red is 10 total (6+4). Now we would put 10 as our numerator. 10/19. We could simplify but 10 and 19 have no common factors. So it would but stay 10/19. Hope this helped

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<u>Confidence interval for population proportion is </u>

<u>sample proportion +/- confidence coefficient*standard error of p </u>

<u>Sample proportion = p = x/n = 180/300 = 0.6 </u>

<u>Confidence coefficient is the critical value of z for 95% confidence level = 1.96 </u>

<u>Standard error of p = sqrt [p*(1-p)/n] </u>

<u>= sqrt [0.6*0.4/300] </u>

<u>= 0.0283 </u>

<u>The CI is </u>

<u>0.6 +/- 1.96*0.0283 </u>

<u>0.6 +/- 0.055 </u>

<u>lower boundary is 0.6-0.055 = 0.545 </u>

<u>upper boundary is 0.6+0.055 = 0.655 </u>

<u>The CI is (0.545, 0.655)</u>

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3 years ago
What is 5 over 8 expressed as a percent?
Vitek1552 [10]

Answer: 62.5%

Step-by-step explanation:

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Mars is 400 million kilometers away, its furthest distance from Earth?
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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
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)\\\\

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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What is $3.850 - $925 ?
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Answer:

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