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3 years ago

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There are four nickels and six dimes in your pocket. You randomly pick a coin out of your pocket and place it on a counter. Then

you randomly pick another coin. Fine the probability of picking a nickel first and a dime second.

1 answer:

Anna71 [15]3 years ago

4 0

The probability of picking a nickel first would be: 4/10
The probability of picking a dime first would be: 6/10

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Write the equation of the line, given the y- and x-intercepts:

sasho [114]

Answer:

y=2x+19

Step-by-step explanation:

Slope: 0-19/-9.5-0=-19/-9.5=2

y-intercept: 19

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(4.23)(1.6) please multiple

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Answer:

The correct answer is 6.768.

Step-by-step explanation:

(4.23)(1.6) = 6.768

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Can someone help me understand this?

ASHA 777 [7]

< means 'less than'. 3.5 is less than 4.5, so A is correct. Since -4.5 is NEGATIVE it is less than 0. As for C, since both numbers are negative, it would be the seemingly 'smaller' number that is actually bigger. It may help to think of the numbers on a number line. In the negatives, the numbers start to go in reverse, so the further from 0 a negative number is, the smaller it is. -4.5 IS LESS than -3.5. All answers are correct.

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On a certain hot summer's day, 477 people used the public swimming pool. The daily prices are 1.50 for children and 2.25 for ad

Akimi4 [234]

Answer:

212 children, and 265 adults

Step-by-step explanation:

To find the number of children and adults, we can set up a systems of equations.

x= number of children

y= number of adults

Equation 1: Price

1.50x+2.25y=914.25

Equation 2: Total number of people

x+y=477

Now, let's solve the equation using substitution.

Rearrange the second equation to solve for one variable.

x+y=477

x=477-y

Now plug x equals into the first equation, and solve for y.

1.50x+2.25y=914.25

1.50(477-y)+2.25y=914.25

715.5-1.50y+2.25y=914.25

715.5+0.75y=914.25

0.75y=198.75

y=265

We just solved for the number of adults. Now let's plug y equals into the second equation to find the number of children.

x+y=477

x+265=477

x=212

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2 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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".

Solution to the problem

The population proportion have the following distribution

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

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.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago