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ANTONII [103]
1 year ago
11

Learn by DoingHere are the directions, grading rubric, and definition of high-quality feedback for the Learn by Doing discussion

board exercises.ContextStudents researching backpack weights gathered data from 45 elementary school children in the 3rd and 5th grades. The variable is "percent of body weight carried in the school backpack." So a child who weighs 60 pounds and carries 9 pounds has a variable value of 15% (9 ÷ 60 = 0.15 = 15%). The American Chiropractic Association (ACA) recommends that children carry no more than 10% of their body weight.PromptWhen we analyze backpack weight as a percentage of body weight, how do 3rd and 5th graders compare? Are children in this study following the ACA recommendation?% of body weight carried in backpack Thirdgraders Fifthgraders0-5% 1 15-10% 6 610-15% 11 415-20% 3 720-25% 0 125-30% 0 230-35% 0 1Totals 21 22Note: Left-hand end-points are included in each bin. So the 2nd bin contains students carrying 5% of their body weight.GradingTo view the grading rubric for this discussion board, click on menu icon (three vertical dots) and then select show rubric. Please note, if viewing the course via the Canvas mobile app the rubric does not appear on this page.Tips for SuccessTo post your initial post, click the "reply" button at the top of the introduction thread below.You are required to reply to two of your peers in this discussion; don't forget to complete this requirement of the activity or you will lose points. Provide high-quality feedback to your peers.
Mathematics
1 answer:
Dahasolnce [82]1 year ago
8 0

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Recommendation from the American Chiropractic Association:

No more than 10\%  of body weight should be borne through kids

Third classification:

More than 10\% of its weight:11 + 3 +0 +0 = 14

ACA Suggestion proportion of children = ( \frac{14}{21}) \times 100 = 66.67\%

Fifth degree:

Over 10\%  of body weight:4 + 7 + 1 + 2 + 1 = 15

ACA Suggestion for children = (\frac{15}{22}) \times 100 = 68.18\%

ACA suggestion doesn't quite follow either majority for third and fifth grades: fifth grades (68.18%) are marginally hyper-third grades (66.67%).

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Sheila has $89 in her bank account. She uses 30% of her money to pay a bill. Chose from the following list:
Veronika [31]

Answer:

The correct option is option c,

Step-by-step explanation:

Sheila has $89 in her bank account original orihinally We want to determine the amount of money which she used to pay percentage a bill. This amount is expressed in terms of percentage of the original amount which she had in her bank.

She uses 30% of her money to pay a bill. Therefore, the amount she used to pay the bill =

30/100 89 = 0.3 ×89

= 26.70.

The correct option is option c,

7 0
2 years ago
A white-tailed deer can sprint at speeds up to 30 miles per hour. American bison can run at speeds up to 3,520 feet per minute.
Paul [167]

3520/5280 = 2/3 of a mile per minute.

2/3 x 60 minutes per hour = 40 miles per hour.

40-30 = 10

Bison runs 40 miles per hour

The bison runs 10 miles an hour faster than the deer

8 0
2 years ago
Read 2 more answers
The state of south dakota has a population of about 814,000 people. Find the population density in people per square miles if th
Gnom [1K]

Remark

Not many per square mile. Sounds like an ideal place to live.

Step One

Find the number of Square Miles.

Area = Height * Width

Area = 380 * 200

Area = 76,000 square miles.

Step Two

Find the Density

<u>Formula</u>

Density = Population / Square Miles

Population = 814,000 people

Square miles = 76000

Solve

Density = 814000/76000 = 10.7 people per square mile.

Remark

Isn't that incredible. It means if all the Area of South Dakota were given to the people living there, each family of 4 would get roughly 1/2 a square mile.

4 0
2 years ago
James is the manager at an entertainment arena that draws an average 7,000 patrons per event. Each ticket taker can process 350
Elina [12.6K]

Answer:

No. James didn't have enough ticket takers to process the average number of patrons hat usually attend the events.

He need to hire 2 more ticket taker (i.e 20 ticket takers) in order to process the average number of patrons hat usually attend the events.

Step-by-step explanation:

Given:

Average amount of patron per event= 7000

Each ticket taker can process = 350

Number of ticket takers hired = 18

We need to find the whether he hire enough to process the average number of patrons that usually attend the events.

Solution:

We will find the Average amount of patron ticket takers can process.

Now we can say that

Average amount of patron ticket takers can process is equal to Each ticket taker can process multiplied by Number of ticket takers hired.

framing in equation form we get;

Average amount of patron ticket takers can process = 350 \times 18 = 6300\ patrons

Hence With the required hires James cannot fully process the patrons usually attending the event.

To find how many ticket takers required we will divide average number of patrons with  Each ticket taker can process.

framing in equation form we get;

ticket takers required = \frac{7000}{350}=20

hence In order t process the the average number of patrons that usually attend the events James will require 20 ticket takers.

6 0
2 years ago
If candies are sold at 3 pcs for php.2.00.how many candies can melai get if she has php20.00?​
Marrrta [24]
If you’re saying that 3 pieces costs 2.00 then the answer is 30. 2x10 is the $20.00 so 3x10 = 30 she can buy
8 0
2 years ago
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