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Ad libitum [116K]
3 years ago
6

In a box-and-whisker plot, the interquartile range is a measure of the spread of the middle half of the data. Find the interquar

tile range for the data set: 10, 3, 7, 6, 9, 12, 13
Mathematics
1 answer:
Bezzdna [24]3 years ago
8 0
In statistics, it is imperative to arrange your data entries from least to greatest. That would be the first step of our solution

3  6  7  9  10  12  13

Interquartile range is the difference of the last quartile to the first quartile. Divide your entire data set into four parts. Each division is called a quartile. Its numerical value is the average of its numbers. However, since our data set contains 7 numbers, which is odd, we write the median (middle value) twice. The quartiles are:

3  6 |  7  9 | 9  10 |  12  13

Q₁= (3+6)/2 = 4.5
Q₂ = (7+9)/2 = 8
Q₃ = (9+10)/2 = 9.5
Q₄ = (12+13)/2 = 12.5

The interquatile range is Q₄ - Q₁ = 12.5 - 4.5 = 8
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mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi
chubhunter [2.5K]

Hello!  

We have the following data:  

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)  

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

a_n = a_1 + (n-1)*r

* second year salary

a_2 = a_1 + (2-1)*300

a_2 = 32000 + 1*300

a_2 = 32000 + 300

\boxed{a_2 = 32300}

* third year salary

a_3 = a_1 + (3-1)*300

a_3 = 32000 + 2*300

a_3 = 32000 + 600

\boxed{a_3 = 32600}

* fourth year salary

a_4 = a_1 + (4-1)*300

a_4 = 32000 + 3*300

a_4 = 32000 + 900

\boxed{a_4 = 32900}

* fifth year salary

a_5 = a_1 + (5-1)*300

a_5 = 32000 + 4*300

a_5 = 32000 + 1200

\boxed{a_5 = 33200}

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

a_6 = a_1 + (6-1)*300

a_6 = 32000 + 5*300

a_6 = 32000 + 1500

\boxed{a_6 = 33500}

* seventh year salary

a_7 = a_1 + (7-1)*300

a_7 = 32000 + 6*300

a_7 = 32000 + 1800

\boxed{a_7 = 33800}

*  eighth year salary

a_8 = a_1 + (8-1)*300

a_8 = 32000 + 7*300

a_8 = 32000 + 2100

\boxed{a_8 = 34100}

* ninth year salary

a_9 = a_1 + (9-1)*300

a_9 = 32000 + 8*300

a_9 = 32000 + 2400

\boxed{a_9 = 34400}

*  tenth year salary

a_{10} = a_1 + (10-1)*300

a_{10} = 32000 + 9*300

a_{10} = 32000 + 2700

\boxed{a_{10} = 34700}

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

*  eleventh year salary

a_{11} = a_1 + (11-1)*300

a_{11} = 32000 + 10*300

a_{11} = 32000 + 3000

\boxed{\boxed{a_{11} = 35000}}\end{array}}\qquad\checkmark

Respuesta:

In the eleventh year of salary he will earn more than 34700, in the case, this value will be 35000

________________________

¡Espero haberte ayudado, saludos... DexteR! =)

7 0
3 years ago
Please help me ASAP and please show me how you did it.
Nadusha1986 [10]

Answer:

2. Find m

sin m˚ = 2/5

sin^-1 (2/5) = m˚

m˚ = 23.57˚

3. Sine law

\frac{8}{sin103}=\frac{5}{sin\alpha }

\frac{5sin103}{8}=sin^{-1}(0.608)=37.5

8 0
2 years ago
Find the focus and directrix of parabola xsquared= 2y
Akimi4 [234]

Answer:

directrix: y = -1/2; focus = (0,1/2)

Step-by-step explanation:

rewrite in standard form:

2y = x^2

factor 4

4 * (y/2) = x^2

rewrite as

4 * 1/2 ( y - 0 ) = (x - 0 )^2

therefore

(h,k) = (0,0) , p = 1/2

directrix is line || to x-axis, distance - p from the center (0,0) y-coordinate (of the parabola)

y = 0 - p

y = 0 - 1/2

y = -1/2

now for the focus!

the focus is distance p from center (0,0) along y-axis --> (0, 0+p)

therefore

(0,0+1/2)

(0,1/2) is the focus!

5 0
2 years ago
Find two consecutive odd integers whose sum is -88
frosja888 [35]
They are -35 and -53
8 0
3 years ago
HELP ME PLEASE <333333333333333
grandymaker [24]
The answer is 15mm^3
7 0
3 years ago
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