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2 years ago

The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament.

1 answer:

4 0

The data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20

<h3>What is mean, median and mode of a data set?</h3>

The mean of the data is the average value of the given data. The mean of the data is the ratio of the sum of all the values of data to the total number of values of data.

The median of the data is the middle value of the data set when it arrange in ascending or descending order.

Mode of a data set is the value, which occurs most times for that data set. The value which has the highest frequency in the given set of data is known as the mode of that data set.

The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament. The table is attached below.

The data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

The mean is-

$\mu=\dfrac{41+41+43+43+45+50+52+53+54+62+63+63+67+75+97}{15}\\\mu=\dfrac{849}{15}\\\mu=56.5$

The median is.

The arranged data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

Here, the middle value is 53. Thus, the median is 53.

The modes from least to greatest are , , and.

Mode is the data which occur most offen. Here the order of modes from least to greatest is,

41 41 43 43 45 50 52 53 54 62 63 63 67 75 97

The range is.

Lowest value in data is 41 and highest is 97. Thus, the range is,

$r=97-41\\r=56$

The interquartile range is.

The third quartile of data is 63 while first quartile is 43. Thus, the interquartile range is.

$IQR=63-23\\IQR=20$

Thus, the data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20.

Learn more about the mean, median and mode here;

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This involves a bit of logic in thinking about what LCM actually means.

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Answer:

$\frac{11}{42}$ or 26.19%

Step-by-step explanation:

If the range of possible numbers is between 12 and 54 then that means that there are 42 (54 - 12) possible numbers that can be made. In this range, there are 11 prime numbers which are the following...

13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53

Therefore, the probability that the number formed is a prime would be $\frac{11}{42}$ which if we divide and multiply the decimal by 100 we get the probability in percentage form.

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Finally, we can see that the probability of this happening is 26.19%

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3 years ago

Whats the correct answer?

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Decide whether the triangles are similar. If so, determine the appropriate expression to solve for x.

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[HELP] For j(x) = <img src="https://tex.z-dn.net/?f=5%5Ex%5E-%5E3" id="TexFormula1" title="5^x^-^3" alt="5^x^-^3" align="absmidd

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The difference quotient of the function that has been presented to us will turn out to be 5.

<h3>How can I calculate the quotient of differences?</h3>

In this step, we wish to determine the difference quotient for the function that was supplied.

To begin, keep in mind that the difference quotient may be calculated by:

Lim h->0 $\frac{f(x+h)-f(x)}{h}$

Now, for the purpose of the function, we need this:

Then we will have:

$\begin{aligned}&\lim _{h \rightarrow 0} \frac{j(x+h)-j(x)}{h} \\&\lim _{h \rightarrow 0} \frac{5 *(x+h)-3-5 * x+3}{h} \\&\lim _{h \rightarrow 0} \frac{5 x+5 h-3-5 x+3}{h} \\&\lim _{h \rightarrow 0} \frac{5 h}{h}=5\end{aligned}$

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