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a_sh-v [17]
2 years ago
11

a group of fitness club members lose a combined total of 28 kilograms in 1 week there are approximately 2.2 pounds in 1 kilogram

assuming the weight loss happened at a constant rate about how many pounds did the group lose each day
Mathematics
2 answers:
ruslelena [56]2 years ago
6 0

Answer:

8.8 pounds per day.

Step-by-step explanation:

We have been given that a group of fitness club members lose a combined total of 28 kilograms in 1 week there are approximately 2.2 pounds in 1 kilogram.

First of all, we will convert the weight loss into pounds by multiplying 28 by 2.2 as shown below:

\text{Weight loss}=28\text{ kg}\times \frac{2.2\text{ pounds}}{\text{ kg}}

\text{Weight loss}=28\times 2.2\text{ pounds}

\text{Weight loss}=61.6\text{ pounds}

Now, we will divide total weight loss by days in week to find weight loss per day.

\text{Weight loss per day}=\frac{61.6\text{ pounds}}{7\text{ days}}

\text{Weight loss per day}=\frac{8.8\text{ pounds}}{\text{ day}}

Therefore, the the group lost 8.8 pounds per day.

notka56 [123]2 years ago
3 0

so heres how you do this

28/7=4 kilograms lost each day

now convert the 4 kilograms to pounds

4 x 2.2 = 8.8 pounds lost each day is your answer

add me as a friend if you like im good at math

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Jenna dances for 3 hours on sunday, 2 hours on Monday and Tuesday, 1 hour on THursday,1.5 hours oon friday, and 2 hours on Satur
Delicious77 [7]

Answer: The average number of hours she danced per day is 1.9 hours (rounded to the nearest tenth)

Step-by-step explanation: We start by calculating how many hours she danced all together which can be derived as follows;

Summation = 3 +2 +2 + 1 + 1.5 + 2 = 11.5

The number of days she danced which is the observed data is 6 days (she did  not dance at all on Wednesday).

The average (or mean) hours she danced each day can be calculated as

Average = ∑x ÷ x

Where ∑x is the summation of all data and x is number of observed data

Average = (3+2+2+1+1.5+2) ÷ 6

Average = 11.5 ÷ 6

Average = 1.9166

Approximately, average hours danced is 1.9 hours (to the nearest tenth)

8 0
2 years ago
Is it possible to form a triangle with side lengths 3.5 3.5, and 9? If so, will it be scalene, isosceles, or equilateral? Or non
Charra [1.4K]

Answer:

It is not possible to draw a triangle with given measurements of 3.5, 3.5, and 9.

Step-by-step explanation:

<em><u>Scalene Triangle</u></em> - All 3 sides have different lengths.

<em><u>I</u></em><em><u>s</u></em><em><u>osceles</u></em><em><u> </u></em><em><u>Triangle</u></em> - 2 sides have equal lengths.

<em><u>Equilateral</u></em><em><u> </u></em><em><u>Triangle</u></em> - All 3 sides have equal lengths.

You must be thinking that it would be Isosceles triangle, but it is not. The measurements you gave is 3.5, 3.5, and 9. Grab a piece of paper, ruler, and a pencil. First draw the length of 9 cm with your pencil and ruler (let us pretend that the measurements are in cm). Then draw 3.5 cm by placing your ruler on the end/start of your 9cm line that you drew before. Then, once again draw a 3.5 cm on the other end of the 9cm line. You will see something like the picture above. You can see that the two sides of the triangle are not intersecting on the top. This means that the triangle formation cannot be made by the given measurements of 3.5, 3.5, and 9.

I hope you understand my answer and this is an easy way to find if, from the given measurements, a triangle is able to be drawn. Thank you !!

4 0
2 years ago
Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.
xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form p(t)=at^2, where a is in ℝ.

2.Let W2 be the set of all polynomials of the form p(t)=t^2+a, where a is in ℝ.

3. Let W3 be the set of all polynomials of the form p(t)=at^2+at, where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

p_1 (t) = at^2, p_2(t)=bt^2.

We must check that p_1+p_2(t) is in W1.

Note that

p_1(t)+p_2(t) = at^2+bt^2  = (a+b)t^2

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say p_1(t) = a(t^2+t), p_2(t)= b(t^2+t). Then

p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 p(t) = at^2and a real scalar b. Then

bp(t) = b(at^2) = (ba)t^2).

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 p(t) = a(t^2+t)and a real scalar b. Then

bp(t) = b(a(t^2+t)) = (ba)(t^2+t).

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.  

6 0
2 years ago
A fruit basket contains oranges and grape fruits. One-third of the oranges and one-fourth of the grapefruits were spoiled. You t
den301095 [7]
So if 1/3 0f the oranges equals 4, than you times it by three to get 3/3, so there were 12 oranges and 28 grapefruits, so the answer is 40
5 0
2 years ago
Marcus solved this problem and found that 190% of 20 is 380. Is he correct? Explain why or why not.
ruslelena [56]

Answer:

no

Step-by-step explanation:

because, 190% as a decimal is 1.9. So 1.9 times 20 is 38. Not 380.

Please mark me brainliest

5 0
2 years ago
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