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Harrizon [31]
3 years ago
8

Answer the statistical measures and create a box and whiskers plot for the following set of data. You may optionally click and d

rag the numbers into numerical order.
4
,
5
,
5
,
7
,
9
,
10
,
10
,
10
,
11
,
12
,
13
,
15
4,5,5,7,9,10,10,10,11,12,13,15
Min: Q1: Med: Q3: Max:
Create the box plot by dragging the lines:
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x
Mathematics
1 answer:
ratelena [41]3 years ago
3 0

sorry

look im trying to finish a challange u need premium if your going toask that big of a quietion my dude/dudet

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A cube with a side length s can be split into three congruent pyramids which formula represents the volume of a square pyramid
Assoli18 [71]
The volume of a triangular pyramid<span> can be found using the </span>formula<span> V = 1/3AH where A = area of the triangle base, and H = height of the </span>pyramid<span> or the distance from the </span>pyramid's<span> base to the apex.</span>
7 0
3 years ago
10+6*(75*50/100)-4/8
sergiy2304 [10]

First, do what is in the parentheses. 75 times 50 is 3,750. Divide it by 100 to get 37.5. 10 + 6(37.5) - 1/2. Do the multiplication to get 10 + 225 - 1/2. Now you get 235 - 1/2. Solve to get 234 1/2.



7 0
3 years ago
Read 2 more answers
Plz help me .. I am really stumped!!!!!!! part of the set for a play is a big triangular piece of plywood. the area of the trian
saul85 [17]
The way to find the area of a triangle is A=\frac{B*H}{2}
if the base is 3 feet longer then the height we can write it like this B=H+3

now we can add it to our area equation
20=\frac{H(H+3)}{2}
20=\frac{ h^{2} +3h}{2}
40=H^{2}+3H
10\frac{1}{3}=H^{2}
h=3.2ft

3 0
3 years ago
Use the disk method or the shell method to find the volumes of the solids generated by revolving the region bounded by the graph
diamong [38]

Answer:

a) 8π

b) 8/3 π

c) 32/5 π

d) 176/15 π

Step-by-step explanation:

Given lines :  y = √x, y = 2, x = 0.

<u>a) The x-axis </u>

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {y.y^2} \, dy  

∴ Vol = 8π

<u>b) The line y = 2  ( using the shell method )</u>

p = 2 - y

h = y^2

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy

     = ( 2π ) * [ 2/3 * y^3  - y^4 / 4 ] ²₀

∴ Vol  = 8/3 π

<u>c) The y-axis  ( using shell method )</u>

h = 2-y  = h = 2 - √x

p = x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

     = (2\pi ) \int\limits^4_0 {x(2-\sqrt{x}  ) } \, dx

     = ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

<u>d) The line x = -1    (using shell method )</u>

p = 1 + x

h = 2√x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

Hence   vol = 176/15 π

attached below is the graphical representation of P and h

3 0
3 years ago
Question 26 pleaseee
LuckyWell [14K]

Answer:

The inequality is 55+10x\leq 105

The greatest length of time Jeremy can rent the jet ski is 5 and Jeremy can rent maximum of 135 minutes.

Step-by-step explanation:

Given: Cost of first hour rent of jet ski is $55

          Cost of each additional 15 minutes of jet ski is $10

          Jeremy can spend no more than $105

Assuming the number of additional 15-minutes increment be "x"

Jeremy´s total spending would be first hour rental fees and additional charges for each 15-minutes of jet ski.

Lets put up an expression for total spending of Jeremy.

\$55+ \$ 10\times x

We also know that Jeremy can not spend more than $105

∴ Putting up the total spending of Jeremy in an inequality.

\$ 55+\$10x\leq \$ 105

Now solving the inequality to find the greatest number of time Jeremy can rent the jet ski,

⇒ \$ 55+\$10x\leq \$105

Subtracting both side by 55

⇒ \$ 10x\leq \$50

Dividing both side by 10

⇒x\leq \frac{50}{10}

∴ x\leq 5

Therefore, Jeremy can rent for 60\ minutes + 5\times 15\\= 60\ minutes + 75= 135\ minutes

Jeremy can rent maximum of 135 minutes.

5 0
3 years ago
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