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rjkz [21]
3 years ago
11

One card is randomly selected from a deck of cards. find the odds in favor of drawing a club.

Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
6 0
->Calculate chance for geting any one card in deck, set a proportion
52(total amount of cards) : 100%(total=100%)=1(any one random card) : X%
X=1.9%
->then multiply percentage of any one card with how many clubs are in the deck, there are 13 clubs
X%•13=24.7% chance of getting clubs
Solution is rounded



Sladkaya [172]3 years ago
4 0

Answer:

The odds in favor is 1:3

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Step-by-step explanation:

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You go out to eat with your family. The subtotal on your bill is $45.90. You must pay a 8% sales tax AND you would like to leave
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Answer:

The answer is 57.83

Step-by-step explanation:

find 8% of 45.90 and add 18% of 45.90 and add them together.

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18%*45.90=8.26

Add, 45.90+8.26+3.67

The final price of your meal is 57.83

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3 years ago
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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

8 0
3 years ago
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