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serg [7]
3 years ago
8

a street that is 116 meters long is covered in snow city workers are using a snow plow to clear the street a tire on the snow pl

ow has to turn 29 times and traveling the length of the street what is the diameter of the tire use the value 3.144 pi
Mathematics
1 answer:
Yuri [45]3 years ago
4 0
So, we know the snow plow tire turned or went around 29 times, namely 29 revolutions, and recall that one revolution is a 2π angle in radians.

since the tire went around 29 times, so it made an angle of 29 * 2π, or 58π.

we know the street is 116 meters long, namely "an arc made by that 58π angle is 116".

\bf \textit{arc's length}\\\\
s=r\theta ~~
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad radians\\
------\\
s=116\\
\theta =58\pi 
\end{cases}\implies 116=r58\pi \implies \cfrac{116}{58\pi }=r
\\\\\\
\cfrac{2}{\pi }=r\qquad \qquad\qquad \qquad  \stackrel{\textit{a \underline{d}iameter is twice the radius}}{2\left( \cfrac{2}{\pi } \right)=d\implies \cfrac{4}{\pi }=d}
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Read 2 more answers
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Vinil7 [7]

Answer:

a)

X[bar]_A= 71.8cm

X[bar]_B= 72cm

b)

M.A.D._A= 8.16cm

M.A.D._B= 5.4cm

c) The data set for Soil A is more variable.

Step-by-step explanation:

Hello!

The data in the stem-and-leaf plots show the heights in cm of Teddy Bear sunflowers grown in two different types of soil (A and B)

To read the data shown in the plots, remember that the first digit of the number is shown in the stem and the second digit is placed in the leaves.

The two data sets, in this case, are arranged in a "back to back" stem plot, which allows you to compare both distributions. In this type of graph, there is one single stem in the middle, shared by both samples, and the leaves are placed to its left and right of it corresponds to the observations of each one of them.

Since the stem is shared by both samples, there can be observations made only in one of the samples. For example in the first row, the stem value is 5, for the "Soil A" sample there is no leaf, this means that there was no plant of 50 ≤ X < 60 but for "Soil B" there was one observation of 59 cm.

X represents the variable of interest, as said before, the height of the Teddy Bear sunflowers.

a) To calculate the average or mean of a data set you have to add all observations of the sample and divide it by the number of observations:

X[bar]= ∑X/n

For soil A

Observations:

61, 61, 62, 65, 70, 71, 75, 81, 82, 90

The total of observations is n_A= 10

∑X_A= 61 + 61 + 62 + 65 + 70 + 71 + 75 + 81 + 82 + 90= 718

X[bar]_A= ∑X_A/n_A= 218/10= 71.8cm

For Soil B

Observations:

59, 63, 69, 70, 72, 73, 76, 77, 78, 83

The total of observations is n_B= 10

∑X_B= 59 + 63 + 69 + 70 + 72 + 73 + 76 + 77 + 78 + 83= 720

X[bar]_B= ∑X_B/n_B= 720/10= 72cm

b) The mean absolute deviation is the average of the absolute deviations of the sample. It is a summary of the sample's dispersion, meaning the greater its value, the greater the sample dispersion.

To calculate the mean absolute dispersion you have to:

1) Find the mean of the sample (done in the previous item)

2) Calculate the absolute difference of each observation and the sample mean |X-X[bar]|

3) Add all absolute differences

4) Divide the summation by the number of observations (sample size,n)

For Soil A

1) X[bar]_A= 71.8cm

2) Absolute differences |X_A-X[bar]_{A}|

|61-71.8|= 10.8

|61-71.8|= 10.8

|62-71.8|= 9.8

|65-71.8|= 6.8

|70-71.8|= 1.8

|71-71.8|= 0.8

|75-71.8|= 3.2

|81-71.8|= 9.2

|82-71.8|= 10.2

|90-71.8|= 18.2

3) Summation of all absolute differences

∑|X_A-X[bar]_A|= 10.8 + 10.8 + 9.8 + 6.8 + 1.8 + 0.8 + 3.2 + 9.2 + 10.2 + 18.2= 81.6

4) M.A.D._A=∑|X_A-X[bar]_A|/n_A= 81.6/10= 8.16cm

For Soil B

1) X[bar]_B= 72cm

2) Absolute differences |X_B-X[bar]_B|

|59-72|= 13

|63-72|= 9

|69-72|= 3

|70-72|= 2

|72-72|= 0

|73-72|= 1

|76-72|= 4

|77-72|= 5

|78-72|= 6

|83-72|= 11

3) Summation of all absolute differences

∑ |X_B-X[bar]_B|= 13 + 9 + 3 + 2 + 0 + 1 + 4 + 5 + 6 + 11= 54

4) M.A.D._B=∑ |X_B-X[bar]_B|/n_B= 54/10= 5.4cm

c)

If you compare both calculated mean absolute deviations, you can see M.A.D._A > M.A.D._B. As said before, the M.A.D. summary of the sample's dispersion. The greater value obtained for "Soil A" indicates this sample has greater variability.

I hope this helps!

7 0
3 years ago
There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con
grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

\square ABCD

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD

\angle BAC = \angle  DCA

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC

and

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED

So, the congruent sides and angles of △ABE and △CDE are:

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD ---- S

\angle BAC = \angle  DCA ---- A

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED or \[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC  --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0
2 years ago
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