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3 years ago

Domain is the set of all input values, while range is the set of all:

Mathematics

1 answer:

Mumz [18]3 years ago

3 0

I am thinking it will be d . Functions

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A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of t

Alexxandr [17]

Answer: A. 0.20

Step-by-step explanation:

Let A be the event of employees needed corrective shoes and B be the event that they needed major dental work .

We are given that : $P(A)=0.08\ ;\ P(B)=0.15\ ;\ P(A\cap B)=0.03$

We know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Then, $P(A\cup B)=0.08+0.15-0.03= 0.20$

Hence, the probability that an employee selected at random will need either corrective shoes or major dental work : $P(A\cup B)= 0.20$

hence, the correct option is (A).

6 0

4 years ago

Because sample means were being tested, the ___ is used to calculate the z-score

Vinvika [58]

<span>Because sample means were being tested, the bell curve is used to calculate the z-score</span>

6 0

3 years ago

Read 2 more answers

Select the correct answer.

OLga [1]

Answer:

The answer is c.

This is because when it rotates a 180 degrees around its center it maps a parallelogram on to itself.

6 0

3 years ago

Which of the following have the correct solution? Check all

Neko [114]

Step-by-step explanation:

To do this, we will substitute the x values to their corresponding equations,

$2x + 5 \\ = 2(7) + 5 \\ = 14 + 5 \\ = 19$

Given 2x + 15 = 19

Calculated 2x + 15 = 19

Given = Calculated, this solution is correct.

$3 + x + 2 -x \\ = 3 + 3 + 2 - 3 \\ = 8 - 3 \\ = 5$

Given 3 + x + 2 - x = 16

Calculated 3 + x + 2 - x = 5

Given not equal to calculated, this solution is incorrect.

$\frac{x + 2}{5} \\ = \frac{8 + 2}{5} \\ = \frac{10}{5} \\ = 2$

Given (x+2)/5 = 2

Calculated (x+2)/5 = 2

Given = Calculated, this solution is correct.

$6 = 2x - 8 \\ 2x - 8 = 6$

$2x - 8 \\ = 2(7) - 8 \\ = 14 - 8 \\ = 6$

Given 2x - 8 = 6

Calculated 2x - 8 = 6

Given = Calculated, this solution is correct.

$14 = \frac{1}{3} x + 5 \\ \frac{1}{3} x + 5 = 14$

$\frac{1}{3} x + 5 \\ = \frac{1}{3}(18) + 5 \\ = \frac{18}{3} + 5 \\ = 6 + 5 \\ = 11$

Given 1/3x + 5 = 14

Calculated 1/3x + 5 = 11

Given not equal to Calculated, this solution is not correct.

Therefore the Correct solutions are 1,3 and 4.

8 0

3 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago