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mario62 [17]
3 years ago
5

Nicholas and Millah are comparing results from a 2-day lab experiment. On Monday, Nicholas’s results showed that 0.917 of the pa

rticles were charged while Millah’s results showed that 85% of the particles were charged. On Tuesday, Nicholas’s results showed that 0.799 of the particles were charged while Millah’s results showed that 93% of the particles were charged. Who on what day had the greatest percentage of charged particles in their experiment?
Mathematics
1 answer:
nalin [4]3 years ago
4 0
Millah Tuesday because they had %93 charged.
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What is the value of (-5)^4
butalik [34]

Answer:

625

Step-by-step explanation:

7 0
3 years ago
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
Devon rented a U Haul truck for $48 for 4 hours plus $13.50 per each hour you go over. At the end of the day, his total bill was
Rasek [7]

Answer:

You need to write an equation to find the answer to this:

h = hours

48 + 13.5h = 112.50               -First you need to subtract 48 from both sides

48 - 48 + 13.5h = 112.50 - 48

13.5h = 64.5         -Now divide both sides by 13.5

13.5h / 13.5 = 64.5 / 13.5

h = 4.77777778

However, I would just round to 5.

I hope this helps!

-Mikayla



3 0
3 years ago
Which equation represents exponential growth?
iogann1982 [59]
The second one is the correct answer .
8 0
2 years ago
5) You and a friend live 4 miles apart. You hear the sound of thunder 18 seconds before your friend hears it. Assuming sound tra
Leokris [45]

Answer:

  see attached

Step-by-step explanation:

At 1100 ft per second for 18 seconds, the sound travels 19,800 ft, or 3.75 miles farther to my friend's house. The set of points that lie 3.75 miles farther from my friend's house than from my house form a hyperbolic curve. This is illustrated by the blue line in the attached graph. (My house is the red dot on the left; my friend's house is the red dot on the right.)

The lightning occurred somewhere on the blue curve.

___

If the lightning occurred on the line between our houses, it was 1/8 mile from my house and 3 7/8 mile from my friend's house. (That's close!)

_____

The formula for the curve in the graph is the distance formula applied to the set of points (x, y). It equates the difference of distance from the two houses to 3.75 miles. If one were to write the equation of the hyperbola in standard form, the equation would look a little different and a restriction would need to be applied so the formula would describe only one branch of the hyperbola.

3 0
3 years ago
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