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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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Ilya [14]

Answer:

The sales will surpass 37 million in about 2013.

Step-by-step explanation:

The answer can be found via solving an exponential inequality as follows:

f(x) = 20(1.05)^x\\20(1.05)^x> 37\\1.05^x> \frac{37}{20}\\x\log 1.05>\log\frac{37}{20}=\log 37 - \log 20\\x>(\log 37 - \log 20)/\log 1.05=12.6\\x>12.6 \,\,\mbox{years}

Any number of years larger than 12.6 will satisfy the inequality, i.e., after more than 12.6 years from year 2000, the annual sales will surpass 37 million. The answer to the exact question will then be: in about 2013 the sales will surpass 37 million.

4 0
3 years ago
Help I’m struggling badly in math
Fynjy0 [20]

Answer:

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Step-by-step explanation:

hope that helped.

5 0
3 years ago
The stopping distance d of an automobile is directly proportional to the square of its speed s.
IRINA_888 [86]

Answer:

192 feet is the stopping distance at 48 mph.

Step-by-step explanation:

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6) Two boats set off from X at the same
lord [1]

Answer:

The distance between the two boats after 2.5 hours is 57 m.

Step-by-step explanation:

The distance covered by each boat after 2.5 hours are:

speed = \frac{distance}{time}

distance = speed x time

For boat A, speed = 14 km/h and time = 2.5 hrs.

So that,

distance = 14 x 2.5

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For boat B, speed = 18 km/h and time = 2.5 hrs.

So that,

distance = 18 x 2.5

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The distance between the two boats after travelling 2.5 hours can be determined by applying the cosine rule. Let the distance be represented by x, so that:

a^{2} = b^{2} + c^{2} - 2bc Cos A

x^{2} = 45^{2} + 35^{2} - 2(45 x 35) Cos 90^{o}

   = 2025 + 1225 - 3150 (0)

   = 3250

x = \sqrt{3250}

  = 57.0088

x = 57.0 m

The distance between the two boats after 2.5 hours is 57 m.

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n200080 [17]
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3 years ago
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