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3 years ago

Mei runs 91.08 miles in 3 weeks. If she runs 6 days each week, what is the average distance she runs each day?

Mathematics

1 answer:

nordsb [41]3 years ago

7 0

Answer:

5.1miles per day

Step-by-step explanation:

Given parameters:

Total distance run in 3 weeks = 91.08miles

Unknown:

Average distance run each day = ?

Solution:

The number of days she ran = 3 x 6 = 18days

Average distance = $\frac{Total distance}{number of days}$

Average distance = $\frac{91.08}{18}$ = 5.1miles per day

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

Tanya [424]

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;

brainly.com/question/14532771

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4 0

2 years ago

28/29 and 30 plz short explanation thanks

zzz [600]

Integers are numbers like ...-3, -2, -1, 0, 1, 2, 3... and so on. So, according to this definition, let's solve this problem:

28. We know that the product of three integers is -3. So, the statement states that we have three factor, so:

a.b.c = -3

The possible values of the factors are:

Factors are: 3, 1 and -1 in which case:

(3)(1)(-1) = -3

Factors are: -3, 1 and 1 in which case:

(-3)(1)(1) = -3

29. In this problem we know two nonzero integers, that is, they cannot be equal to zero. Thus, they must have different sign, that is, one of them must be positive and the other negative. For instance:

-9 and 2

(-9)(2) = -18, So -18 is less than -9 and 2

Another example:

1 and -8

(1)(-8) = -8, So -8 is equal to -8 and -8 is less than 1.

30. The sign of the product of two integers with the same sign is positive, that is:

(+)(+) = (+) and

(-)(-) = (+)

So, the sign of the product of three integers with the same sign are two options:

<u>Option 1.</u> The three integers are positive in which case:

(+)(+)(+) = (+), So the sign of the product is positive

<u>Option 2.</u> The three integers are negative in which case:

(-)(-)(-) = (-), So the sign of the product is negative.

This option two gives us that result because:

(-)(-) = (+) and (+)(-) = (-)

5 0

3 years ago

Which is the difference of 21 1/4 - 18 2/4? A:2 1/4. B:2 2/4. C:2 3/4. D:3 1/4.

seropon [69]

Answer:

2 3/4 C

Step-by-step explanation:

$(21+\frac{1}{4})-(18+\frac{2}{4})\\=21-18+\frac{1}{4} -\frac{2}{4} \\=3-\frac{1}{4} \\=\frac{12-1}{4} \\=\frac{11}{4} \\=2\frac{3}{4} \\or\\21\frac{1}{4} -18\frac{2}{4} \\=\frac{85}{4} -\frac{74}{4} \\=\frac{85-74}{4} \\=\frac{11}{4} \\=2\frac{3}{4}$