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2 years ago

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the value of a collectible coin can be represented by the equation y=2x+9.74 where x represents the number of years that consuel

lo has owend the coin y represents the total value in dollars of the coin. what was the value of the coin when consuello originally purchased it?

1 answer:

Serggg [28]2 years ago

4 0

Answer:

The answer is b ($9.74)

Step-by-step explanation:

I got it right (:

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State, with an explanation, whether the mean, median, or mode gives the best description of the following average. The average

erma4kov [3.2K]

Answer:

The mean is the better method.

Step-by-step explanation:

The best way to meassure the average height is throught mean. The mean of a sample is the average of that sample's height, and it will be a good estimate for the population's average height.

The mode just finds the most frequent height. Even tough the most frequent height will influence the average height, knowing only what height is the most frequent one doesnt give you enough informtation about how the height is centrally distributed.

As for the median, it is fine to use the median of a sample to estimate the median of the population, but if you use the median to estimate the average height you may have a few issues. For example, if you include babies in your population, the babies will push the average height down a lot and they are far below te median height. This, as a result, will give you a median height of a sample way above the average height of the population, becuase median just weights every person's height the same, while average will weight extreme values more, in the sense that a small proportion of extreme values can push the average far from the median.

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1 year ago

A 2-column table has 4 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1. The second column is label

tangare [24]

Answer:

5

Step-by-step explanation:

take 0.5 divided by 5 and it gives you your next number on the chart:)

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1 year ago

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In the diagram, m ∥ n and m and n are cut by transversal p. Which angle pairs are congruent? Check all that apply. ∠1 and ∠5 ∠3

Gelneren [198K]

The picture in the attached figure

we know that

a) <span>∠1 and ∠5
are congruent-----> by corresponding angles

b) </span>∠3 and ∠8------> supplementary angles (∠3 + ∠8=180°)
<span>are not congruent

c)</span>∠6 and ∠4------> supplementary angles (∠6 + ∠4=180°)
are not congruent

d) ∠8 and ∠2------> supplementary angles (∠8 + ∠2=180°)
are not congruent

e) ∠4 and ∠7-----> supplementary angles (∠4 + ∠7=180°)
are not congruent

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2 years ago

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An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

$\large t_{0.05}=2.365$

Now we can compute our confidence interval

$\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546$

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

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2 years ago

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2 years ago

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