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3 years ago

6

A test contains two multiple-choice questions. If a student makes a random guess to answer each question. (Hint: Consider two ou

tcomes for each question—either the answer is correct or it is wrong.) 2.2.1 How many outcomes are possible?

1 answer:

Maslowich3 years ago

5 0

Answer: 2

Step-by-step explanation: he could ethier get t right or get it wrong

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A campsite charges a flat fee per vehicle and a price per camper for each evening. The table shows the number of campers, 2, at

11Alexandr11 [23.1K]

Answer:

dollars per camper

Step-by-step explanation:

If the number of campers is considered as domain of the function, then the amount of charge is the range of this function

The rate of change is the <u>change in range</u> per <u>change in domain</u>, therefore it can be represented by:

dollars per camper

6 0

3 years ago

18 - (-6x) - 3 + 8x<br><br> simplify the expression

EastWind [94]

Answer:

16+13x

Step-by-step explanation:

6 0

3 years ago

Change to mixed number <br><br> 42/8

Llana [10]

Answer:

42/8= 5 2/8

explanation:

8*5=40 42-40=2

therefore

42/8= 5 2/8

4 0

2 years ago

Read 2 more answers

Doss [256]

Answer:

First point: (1,0)

Second Point: (2,1)

Third Point: (3,2)

Fourth Point: (8,7)

Step-by-step explanation:

For the first two points, you have to plot the two pairs of coordinates that are given. After this, you have to plug the x values into the equation to get the missing y values.

For example: The third point

X= 3

Y= X - 1

Y= 3 - 1

Y=2

So this means that the third coordinate is (3,2)

3 0

3 years ago

Consider the function f(x) = x^3 - 7x^2 - 2x + 14

mars1129 [50]

Answer:

There are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14$

To get the zeros of $f(x)$ , set $y$ or $f(x) =0$

so

$0=x^3-7x^2-2x+14$

as

$x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)$

so

$\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0$

Using the zero factor principle:

$\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0$

solving

$x-7=0$

$x=7$

$x+\sqrt{2}=0$

$x=-\sqrt{2}$

$x-\sqrt{2}=0$

$x=\sqrt{2}$

Therefore, there are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

8 0

3 years ago