The following box plot shows the number of years during which 20 schools have participated in an interschool swimming meet:

How to get the greatest common of 45 and 81

erastova [34]

Find the factors and get the ones common to both of them

45=3 times 3 times 5
81=3 times 3 times 3 times 3
so the factors common to both are 3 times 3 which is 9

5 0

3 years ago

Read 2 more answers

What is 4r + 3r + 7y + 5r - 2y - 6r ?<br><br>A. 5r + 5y<br><br>B. 29r - 27y<br><br>C. 27ry

Sonja [21]

Step-by-step explanation:

It's 6r+5y. But it's not in the option

3 0

3 years ago

A biologist tags 25 birds with a band on their foot and lets them go back into the wild. The next year 100 total birds are captu

mestny [16]

Answer:

We can estimate a population of 500 birds.

Step-by-step explanation:

We can estimate the total bird population using a rule of three with the information given:

If we find 5 birds with the foot band among 100 birds captured, how many birds there are in total if inicially we put foot bands in 25 birds?

5 birds with foot band -> 100 birds captured

25 birds with foot band -> X total birds

X * 5 = 25 * 100

X = 25 * 100 / 5 = 500 birds

We can estimate a population of 500 birds.

7 0

3 years ago

anna is in acave 42 feet below the cave entrance. she descends 17 feet . then ascends 21 feet. find her new position relative to

tatyana61 [14]

The answer is Negative 38

3 0

3 years ago

Although companies would like consumers to believe that Identity theft protection is an

stiks02 [169]

The obtained answers for the given frequency distribution are:

(a) The formula for the mean in sigma notation is $\bar x =\frac{1}{n} \sum X_i$ where n is the number of observations; $X_i$ are the n observations.

The mean for the given monthly plan price is $16.1.

(b) The frequency distribution for given data is {$9.99 - 2; $10 - 5; $12 - 1; $12.75 - 2; $14.99 - 6; $20 - 4; $25 - 5}

(c) The formula for the mean using the frequency distribution table is $\bar x = \frac{1}{N}\sum f_ix_i$ where $N =\sum f_i$ and on applying this formula for the given data, the mean is $16.1.

(d) The median for the given data is $m_e = 14.99$ , and the mode for the given data is $14.99

<h3>What are the mean, median, and mode for a frequency distribution?</h3>

The frequency distribution has sample observations $x_i$ and frequencies $f_i$ .

Then, the mean is calculated by

$\bar x = \frac{1}{N}\sum f_ix_i$

Where $N =\sum f_i$ (Sum of frequencies)

The median is calculated by

$m_e=\left \{ {{x_{k}} \ if \ n = 2k+1 \atop {\frac{x_{k}+x_{k+1}}{2}} \ if \ n =2k} \right.$

The mode is calculated by

Mode = highest frequency value

<h3>Calculation:</h3>

The given list of data is

{$14.99, $12.75, $14.99, $14.99, $9.99, $25, $25, $10, $14.99, $10, $20, $10, $20, $14.99, $10, $25, $20, $12, $14.99, $25, $25, $20, $12.75, $10, $9.99}

(a) Formula for the mean using sigma notation and use it to calculate the mean:

The formula for the mean is

$\bar x =\frac{1}{n} \sum X_i$

Where n = 25; $X_i$ - n observations

On substituting,

Mean $\bar x$

=1/25(14.99+12.75+14.99+14.99+9.99+25+25+10+14.99+10+20+10+20+14.99+10+25+20+12+14.99+25+25+20+12.75+10+9.99)

= 1/25(402.42)

= 16.09 ≅ 16.1

(b) Constructing a frequency distribution for the data:

Cost - frequency - cumulative frequency

$9.99 - 2 - 2

$10 - 5 - 7

$12 - 1 - 8

$12.75 - 2 - 10

$14.99 - 6 - 16

$20 - 4 - 20

$25 - 5 - 25

Sum of frequencies N = 25;

(c) Using frequency distribution, calculating the mean:

The formula for finding the mean using frequency distribution is

$\bar x = \frac{1}{N}\sum f_ix_i$

Where N = 25;

On substituting,

$\bar x$ <em> </em>= 1/25 (2 × 9.99 + 5 × 10 + 1 × 12 + 2 × 12.75 + 6 × 14.99 + 4 × 20 + 5 × 25)

= 1/25 (402.42)

= 16.09 ≅ 16.1

Therefore, the mean is the same as the mean obtained in option (a).

(d) Calculating the median and the mode:

Since N = 25(odd) i.e., 2· 12 + 1; k = (12 + 1)th term = 13th term

So, the median $m_e$ = 14.99. (frequency at 13th term)

Since the highest frequency is 6 occurred by the cost is $14.99,

Mode = 14.99

Learn more about frequency distribution here:

brainly.com/question/27820465

#SPJ9

7 0

2 years ago