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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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Alexis has a rectangular piece of red paper that is 4 centimeters wide. It’s length is twice it’s width. She glued a rectangular
Montano1993 [528]

for the rectangular red paper :

w₁ = width of the red paper = 4 cm

L₁ = length of red paper = 2 w = 2 x 4 = 8 cm

A₁ = area of the red paper = L₁ w₁ = 8 x 4 = 32 cm²


for the rectangular blue paper :

w₂ = width of the blue paper = 3 cm

L₂ = length of blue paper = 7 cm

A₂ = area of the blue paper = L₂ w₂ = 7 x 3 = 21 cm²

area visible is given as

A = A₁ - A₂ = 32 cm² - 21 cm² = 11 cm²

so 11 square units of red paper will be visible on top



5 0
2 years ago
Which statement is true about the graphed function
natka813 [3]

From -∞ to -4 the blue line is above the X axis which means it is >0

The blue line is negative between -4 and -3


This would make the correct answer: F(x) > 0 over the interval (-∞,-4)

8 0
2 years ago
Read 2 more answers
Rationalize the denominator of $\frac{2}{3\sqrt{5} + 2\sqrt{11}}$ and write your answer in the form $\displaystyle \frac{A\sqrt{
vlada-n [284]

Answer:

<h2>19</h2>

Step-by-step explanation:

Given the surdic expression \frac{2}{3\sqrt{5} + 2\sqrt{11}}\\, to rationalize the expression, we will have to multiply the numerator and denominator of the expression by the conjugate of the denominator as shown;

= \frac{2}{3\sqrt{5} + 2\sqrt{11}} * \frac{3\sqrt{5} - 2\sqrt{11}}{3\sqrt{5} - 2\sqrt{11}}\\\\= \frac{2(3\sqrt{5} - 2\sqrt{11})}{(3\sqrt{5} + 2\sqrt{11})(3\sqrt{5} - 2\sqrt{11})}\\\\= \frac{6\sqrt{5} - 4\sqrt{11}  }{9\sqrt{25}+6\sqrt{55}- 6\sqrt{55}-4\sqrt{121}  } \\\\=  \frac{6\sqrt{5} - 4\sqrt{11}  }{9(5)-4(11)  }\\\\=  \frac{6\sqrt{5} - 4\sqrt{11}  }{45-44  }\\\\= \frac{6\sqrt{5} - 4\sqrt{11}  }{1}

Comparing the result \frac{6\sqrt{5} - 4\sqrt{11}  }{1} with the expression \frac{A\sqrt{B} + C\sqrt{D}}{E}, it can be seen that A = 6, B = 5, C = -4, D = 11 and E = 1

A+B+C+D+E = 6+5+(-4)+11+1

A+B+C+D+E = 11-4+12

A+B+C+D+E = 19

Hence the value of A+B+C+D+E is 19

4 0
2 years ago
max and some friends shared the cost of a meal. The meal cost $51 and each person contributes $17. How many people share the cos
stiv31 [10]

Answer:

3 people

Step-by-step explanation:

$51/$17 = 3 people

- We know to start with $51 dollars because that is the total price of the meal.

- We also know each meal costs $17.

- Next, ask youself what we are trying to find? We are trying to find how many people are eating.

- Therefore to find that we need to take our starting amount $51 divided by $17 because that is the cost of 1 meal.

- $51/$17 = 3 people (whole cost/individual cost = # of people)

- To check this take 3 x 17 = 51 (Now we know it is correct)

- If you would like a further explanation please let me know.

5 0
2 years ago
In a network of 40 computers, 5 hold a copy of a particular file. Suppose that 7 computers at random fail. Let F denote the numb
anzhelika [568]

Answer:

a. E(F)=0.875

b. 99.9976%

c. P(X=2)=0.1683

Step-by-step explanation:

a. We notice that this is a binomial distribution with the probability of success;

p=\frac{5}{40}=0.125

#We are given the sample size, n=7. The Expected value is calculated as:

E(X)=np\\\\E(F)=np , n=7, p=0.125\\\\E(F)=7\times 0.125\\\\=0.875

Hence the expectation, E(F)=0.875

b. To calculate the probability of the range of F, we need to calculate all possible outcomes of F in the given sample;

P(X\leq 5)=1-P(X=6)-P(X=7)\\\\=1-{7\choose 6}(0.125)^6(1-0.125)^1-{7\choose 7}(0.125)^7(1-0.125)^0\\\\=1-0.000023365-0.000000476\\\\=0.999976158\\\\=99.9976\%

Hence, the range of F is 99.9976%

c. The probability that F=2 is calculated  using the binomial distribution function as:

P(X=2)={7\choose 2}(0.125)^2(1-0.125)^5\\\\=0.1683

Hence, the probability of F=2 is 0.1683

3 0
2 years ago
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