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zalisa [80]
2 years ago
12

Currently, an employee can assemble on average, x amount of masks per hour. An INDLS student makes twice as many N95 Masks as th

e average employee (2x). A public school student makes 1 less than average (x-1). The product of the two employees’ output equals 12 masks per hour.
Using complete sentences:

1. State the function f(x)

2. Show the function in factored form

3. Find the zeros

4. Which zero or zeros make since in this context?

5. Explain each item and show all work.
Mathematics
1 answer:
PtichkaEL [24]2 years ago
8 0
The correct answers is 45 masks per hour
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Natalie's group brought in pizza but is buying drinks at the concession stand medium sodas are a dollar and $.25 how many medium
Phantasy [73]

Keywords:

<em>Medium sodas, buy, dollars, divide </em>

For this case we must find the amount of medium sodas that Natalie's group can buy, taking into account that they have 20 dollars and that each medium soda costs 1.25 dollars. To solve, we must divide:

Let "x" be the number of medium sodas you can buy, then:

x = \frac {20} {1.25}\\x = 16

So, Natalie's group can buy 16 medium sodas with 20 dollars

Answer:

16 medium sodas

8 0
3 years ago
The lengths of lumber a machine cuts are normally distributed with a mean of 106 inches and a standard deviation of 0.3 inch. ​(
Vinvika [58]

Answer:

a) P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)

And we can find this probability with the complement rule:

P(z>0.37)=1-P(z

b) z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}=  2.431

And if we use the z score we got:

P(z>2.431) =1-P(z

Step-by-step explanation:

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

X \sim N(106,0.3)  

Where \mu=106 and \sigma=0.3

Part a

We are interested on this probability

P(X>106.11)

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And using this formula we got:

P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)

And we can find this probability with the complement rule:

P(z>0.37)=1-P(z

Part b

For this case we select a sample of n =44 and the new z score formula is given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score we got:

z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}=  2.431

And if we use the z score we got:

P(z>2.431) =1-P(z

6 0
2 years ago
Find the product and express your answer in scientific notation. (1.3 x 10-6)(4 x 1010)
sveticcg [70]
(1.3\cdot10^{-6})(4\cdot10^{10})=(1.3\cdot4)(10^{-6}\cdot10^{10})=5.2\cdot10^{-6+10}\\\\=\boxed{5.2\cdot10^4}
3 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
Geometry help acute obtuse or right will give brainliest
ira [324]
If I'm reading this right, than all 9 angles are acute?
7 0
2 years ago
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