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jonny [76]
3 years ago
10

The physical education instructor asked each student to do a total of 36 pull-ups and push-ups in 1 minute. The instructor wante

d students to do 8 times as many push-ups as pull-ups. Write a system of linear equations that represents this situation. How many pull-ups and push-ups were required in 1 minute?
Mathematics
1 answer:
NeTakaya3 years ago
8 0

Answer: Pull-ups = 4 and push up = 32

Step-by-step explanation:

Let x represents number of pull ups in one minute.

And, y represents number of push ups in one minute.

Therefore, According to the question,

Each student to do a total of 36 pull-ups and push-ups in 1 minute.

⇒ x+y= 36

But, again from equation, The instructor wanted students to do 8 times as many push-ups as pull-ups.

Therefore, 8x = y

Thus, required system of linear equations that represents this situation is,

x+y= 36 and 8x = y , where x is the number of pull ups and y is the number of push ups in one minute.

By solving these two equations we get, x=4 and y= 32

That is, In one minute 4 pull ups and 32 push ups are required.




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ludmilkaskok [199]

Answer:

a=21

AB=23.9

BC=19.5

CD=(23.9)

AD=19.5

Step-by-step explanation:

Each pair of parallel sides are equal in a parallelogram. Sense both pairs use the same variable we can just choose the easiest to solve which would be:

a-1.5=19.5

add 1.5 to both sides

a=21

now substitute 21 in for a in each equation.

AB=CD so

AB=23.9

BC=AD so

BC=19.5

CD=(21)+2.9 so

CD=(23.9)

AD is given so

AD=19.5

4 0
2 years ago
Please help and explain. i’m confused.
kondor19780726 [428]

Answer:  628

Step-by-step explanation:

3.14*100*6/3= 628

8 0
3 years ago
For carolina birthday her mom took her and 4 friendse to a waterpark.Carolina mom paíd 40 for 5 students tickets.What was the pr
Vinvika [58]

40 dollars per 5 students=9 dollars per 1 student

divide both sides by 5.

7 0
3 years ago
Holly spends 7 3/5 hours in school each day. Her lunch is 1/2 hour long, and she spends 7/10 of a hour switching classes. The re
Ira Lisetskai [31]
<span>
y = 7 + 3/5
y = 35/5 + 3/5
y = 38/5
y = 2*(38/5)
y = 76/10
---
lunch time:
z = 1/2
z = 5*(1/2)
z = 5/10
---
time switching classes:
w = 7/10
---
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (76/10 - 5/10 - 7/10)/6
x = (76 - 5 - 7)/(10*6)
x = (64)/(10*6)
x = (2*2*2*2*2*2)/(2*5*2*3)
x = (2*2*2*2)/(5*3)
x = 16/15

x = 1.0666666666
---
check:
y = 7 + 3/5
y = 7.6
z = 1/2
z = 0.5
w = 7/10
w = 0.7
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (7.6 - 0.5 - 0.7)/6
x = 1.0666666666

answer:
 1.07 hours</span>
5 0
3 years ago
A professor went to a website for rating professors and looked up the quality rating and also the "easiness" of the six full-tim
Marat540 [252]

We are asked to determine the correlation factor "r" of the given table. To do that we will first label the column for "Quality" as "x" and the column for "Easiness" as "y". Like this:

Now, we create another column with the product of "x" and "y". Like this:

Now, we will add another column with the squares of the values of "x". Like this:

Now, we add another column with the squares of the values of "y":

Now, we sum the values on each of the columns:

Now, to get the correlation factor we use the following formula:

r=\frac{n\Sigma xy-\Sigma x\Sigma y}{\sqrt{(n\Sigma x^2-(\Sigma x)^2)(n\Sigma y^2-(\Sigma y)^2)}}

Where:

\begin{gathered} \Sigma xy=\text{ sum of the column of xy} \\ \Sigma x=\text{ sum of the column x} \\ \Sigma y=\text{ sum of the column y} \\ \Sigma x^2=\text{ sum of the column x\textasciicircum2} \\ \Sigma y^2=\text{ sum of the column y\textasciicircum2} \\ n=\text{ number of rows} \end{gathered}

Now we substitute the values, we get:

r=\frac{\left(6)(70.56)-(25.2)(16.4\right)}{\sqrt{((6)(107.12)-(25.2)^2)((6)(47.82)-(16.4)^2)}}

Solving the operations:

r=0.858

Therefore, the correlation factor is 0.858. If the correlation factor approaches the values of +1, this means that there is a strong linear correlation between the variables "x" and "y" and this correlation tends to be with a positive slope.

7 0
1 year ago
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