In how many ways can you receive four cards of the same face value and one card from the other 48 available cards?

1 answer:

5 0

There are 13 sets of 4 cards in a deck of card. So, there are 13 ways to get four cards of the same face value. The next card you pick will always be a different face value. So, essentially, you have 13 x 48 combinations of 5 cards that you can pick from the deck where you have 4 cards of the same face value and 1 card of a different face value.

13 x 48 = 624.

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In 2011, a train carried 8% more passengers than in 2010. In 2012 , it carried 8% more passengers than in 2011. Find the percent

konstantin123 [22]

Answer:

there was a 16% increase since 2010 to 2012

Step-by-step explanation:

lets say the train had 100 people on it in 2010 and in 2011 it had an 8% increase which then brings you to 8 plus 100 and you get 108.

Now in 2012 there was an increase 8%, so you will add 8 to your previous answer 108 and receive 116. So now there was 16 more passengers in 2012 than in 2010. Transfer the 16 passengers into 16% and you now have your answer.

5 0

2 years ago

Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.ca

Phoenix [80]

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

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".

$p$ represent the real population proportion of interest

$\hat p =0.09$ represent the estimated proportion for the sample

n=800 is the sample size required

$z$ represent the critical value for the margin of error

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.