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AveGali [126]
3 years ago
15

A grocery store manager received two shipments of oranges. According to the distributor of the oranges, the oranges in both ship

ments were equivalent because all of the oranges ranged in weight from 140 to 190 grams. The grocery store manager decided to weigh a sample of 25 oranges from each shipment to compare the weights of the oranges. The line plots show the manager's results
Based on these samples, which statement is true about the two shipments of oranges?


The two shipments have about the same mean weight but a different range of weights.

The two shipments have a different mean weight and a different range of weights.

The two shipments have about the same mean weight and the same range of weights.

The two shipments have the same range of weights but a different mean weight.

Mathematics
2 answers:
Volgvan3 years ago
7 0
The answer is
 <span>The two shipments have the same range of weights but a different mean weight.
Hope i helped </span>
Mashutka [201]3 years ago
3 0

We have been given two line plots which represents the data weight of 25 oranges of two shipments created by  grocery store manager.

We have to find which of the given statements are true. To answer our problem we have to find the range and mean of the given data.

The range of a data set can be calculated  subtracting the lowest value of data set from highest value of data set. By subtracting lowest value from highest value of both of our data set we get

190-140=50

Since both data sets have same highest and lowest values our range is same that is equal to 50.

Now we will find mean of both data sets. Let us start with shipment 1 data.

Mean of shipment 1 = \frac{140+145+150+2(155)+2(160)+3(165)+4(170)+5(175)+3(180)+2(185)+190}{25}\\&#10;=\frac{4215}{25}\\&#10;=\frac{843}{5}=168.6

Now we will find mean of shipment 2.

Mean of shipment 2 =\frac{140+3(145)+5(150)+4(155)+3(160)+2(165)+2(170)+2(175)+180+185+190}{25}&#10;=\frac{4000}{25}&#10;=160

After calculating mean of both data sets we get mean of both data sets are different. Now look at the provided options which is equivalent to our answer.

The last option is correct because we got the same result in our answer that two shipments have the same range of weights but a different mean weight.

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sveta [45]

Another way of finding Surface Area is

  1. SA= 2LW+2LH+2HW

L= Length

W= Width

H= Height

the answer, and explained, would be this

2. SA= 2(12)(10)+2(12)(16)+2 (10)(16)

(<em>multiply</em>)

3. SA= 240 + 384 + 320

(<em>add</em>)

<em>The</em><em> </em><em>answer</em><em> </em><em>is</em><em>.</em><em>.</em><em> </em>

<em><u>SA</u></em><em><u>=</u></em><em><u>944ft3</u></em><em><u> </u></em>

7 0
3 years ago
Find the slope of the line passing through the points (-4,7) and (2,-9)
Ira Lisetskai [31]

Answer: The slope is -8/3

Step-by-step explanation: Use the slope formula y2-y1/x2-x1

so -9-7/2-(-4) = -16/6 which can be divided by 2 to get -8/3

7 0
2 years ago
Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

5 0
3 years ago
The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If
stira [4]
The is the concept of algebra, the weight of the spring can be modeled by the equation;
w=0.9sqrt E
where;
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thus the weight of the spring when the energy is 12 j will be:
w=0.9sqrt12
w=3.18 g
4 0
4 years ago
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lilavasa [31]

Answer:

1

Step-by-step explanation:

Only 1, because rhombuses have every side length the same.

8 0
3 years ago
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