All categories

3 years ago

Find the range of the data in the box plot below.

Mathematics

1 answer:

mezya [45]3 years ago

8 0

Answer:

The range of the data is: 4.5 dollars

Step-by-step explanation:

From the diagram,

Smallest value = 5

First Quartile = 5.5

Largest value = 9.5

Third Quartile = 7.5

Median = 6

The range is given by subtracting the smallest value from the largest value:

$9.5 - 5 = 4.5$

Thus, the range of the data is: 4.5 dollars

You might be interested in

How to use words to describe the two ways you can think about 8 X 5 = 40 as a comparison?

stiks02 [169]

Answer:

opo

Step-by-step explanation:

tulog kana po, goodnight

7 0

2 years ago

Read 2 more answers

Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Sea-hawks when they won Su

weeeeeb [17]

Answer:

No. These measures show a thinner team of NFL players according to the mean, variance, standard deviation, and quartiles.

Step-by-step explanation:

1) The measures of variation, namely The Range, Variance, Quartiles, Interquartiles, Sum of Squares, etc. shows us how the data are dispersed.

The Range Δ is calculated:

$\Delta =305-189=116$ Maximum value - Minimum value for weight

Mean:

$\bar{x}=\frac{189+254+235+225+190+305+202+190+252+305}{11}\approx 231.09 \:pounds$

Variance:

$s^{2}=\frac{\sum_{i=1}^{n} (x_i-\bar{x})^{2}}{n-1}\Rightarrow s=1923.69\\$

The Standard Deviation of the sample

$s=\sqrt{s^{^{2}}}\Rightarrow s\approx 43.86\\$

2) Since there is no preceding exercise, the comparison was made to a recent study in which a NFL player average weight is about 245 pounds (average),

Since 25% of this list are player whose weight is 192.5 lbs and 50% (2nd Quartile) =225 lbs , finally only at the 3rd Quartile we have players above the regular NFL average with 253. This, added with the other data, allow us to say that this list is not a typical of all NFL players.

4 0

3 years ago

Nora got an allowance of 50¢ each weekday and 75¢ each weekend altogether how much do they get in:

Furkat [3]

A) > since it is 50c per weekday and 75c each weekend assuming it allows for the 2 days each saturday/sunday.
50c * 5 = $2.50 since there are 5 days in weekdays
75c * 2 = $1.50 since there are 2 days in the weekend
Add $2.50 and $1.50 to get $4.00

b) For 3 school days we know it is a weekday on the school week.
So perform 50c * 3 which gives us <span>$1.50

</span>c) 12 days off from school is 10 weekdays and 1 weekend or 2 days of 75c
So now just do 50c * 10 which is $5.00 and 75c * 2 which is $1.50
Add $5.00 and $1.50 and we get $6.50

d) 4 weeks = 20 weekdays since 5 *4 = 20 and 8 days in each weekend since 2 * 4 = 8
Now that we have the amount of weekdays and weekend days we can multiply.
50c * 20 = $10.00
75c * 4 = $3.00
Add $10.00 and $3.00 to get $13.00 for 4 weeks.

e) We have 1 day of the weekend and 2 weekdays here.
50c * 2 = $1.00
75c * 1 = 75c
<span>$1.00 + 75c = $1.75 in those 3 days listed.
</span>
Add all these together to get your total value.
$4.00 + $1.50 + $6.50 + $13.00 + $1.75 = $26.75

7 0

3 years ago

Find the length of the missing side of the right triangle given the area of the two triangles find the length of the sides of th

irina [24]

From here, it looks like all three sides of the right triangle are missing. I don't see any of them in the question.

3 0

3 years ago

Question is in the picture pls help.

Usimov [2.4K]

Answer:

c. is the answer

5 0

3 years ago

Read 2 more answers