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SashulF [63]
2 years ago
10

(please help) There are 251 students from Mason Middle School going on a feild trip. The students will be riding on buses that h

old 52 students each. How many buses will be needed and how many empty seats will there be? Explain your answer.
Mathematics
2 answers:
Grace [21]2 years ago
7 0
There would be 5 buses and 9 extra seats because 5 x 52 is 260 and you need 251 and there would be 9 extra seats
ExtremeBDS [4]2 years ago
3 0

Answer:

5 buses and 9 empty seats

Step-by-step explanation:

What I did im not sure if its right but i did 251/52 and i got 4.8 which rounds to 5 buses. Then 5*52 would equal 260, then 260-251=9

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Pluto is looking into joining a gym. He has a budget of $68 per month. The gym he wants to join has fees based on the number of
Artist 52 [7]
7(10 visits) ≤ 68
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Not Viable


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Viable


The first solution is not viable because 10 visits will give a total cost of $70 per month, which is $2 over his monthly budget of $68.

The second solution is viable. He could visit the gym a maximum of 9 times per month for a total of $63 a month and that will be under his $68 monthly budget.

Hope this helps! :)
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3 years ago
What is the mean of this data?<br><br> 12, 18, 11, 25, 38, 22
yanalaym [24]
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3 years ago
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4 0
3 years ago
Read 2 more answers
3. Let A, B, C be sets and let ????: ???? → ???? and ????: ???? → ????be two functions. Prove or find a counterexample to each o
Fiesta28 [93]

Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not  injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true  that f must be surjective? Is it true that g must be surjective? Justify your answers with either a  counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not  injective (since both 2, 3 7→ 4) but g ◦ f is injective.  Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2  . Then g is not  injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t  specify the domain and codomain though. Note that the codomain of f must equal the domain of

g for g ◦ f to make sense.

(2) Answer

Solution: There are two questions in this problem.

Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}.  Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that  g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective.  Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we  must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the  c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).

Then g(b) = g(f(a)) = c and we have found our desired b.  Remark: It is good to compare the answer to this problem to the answer to the two problems

on the previous page.  The part of this problem most groups had the most issue with was the second. Everyone should  be comfortable with carefully proving a function is surjective by the time we get to the midterm.

3 0
3 years ago
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