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Zigmanuir [339]

2 years ago

7

Brandon wants to create a probability experiment with a coin toss. He plans to conduct eight trials in his experiment. He predic

ts that half of the trials will result in tails.
A) Is his prediction an example of relative frequency or theoretical probability and why?
B) Which of the two graphs could not be a graph that represents the results of the 8 trials?
SHOW ALL YOUR WORK

1 answer:

trasher [3.6K]2 years ago

5 0

i need the answer i need help plz give the answear and the work hot to do it plz

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*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago

A truck is rented at $40 per day plus a charge per mile of use. The truck traveled 15 miles in one day, and the total was $115.

Sergio039 [100]

Answer: The equation you will use would be 40x + 15x =115. you're welcome.

4 0

2 years ago

I do not know what you do on problem 14

Elan Coil [88]

Distance d is directly proportional to time t that he rides.

So d = kt where k is the constant of proportionality

Here k = his speed = 20 mph

So the required equation is d = 20t

5 0

3 years ago

Write an equation to match each graph:

morpeh [17]

Write an equation to match each graph.

Solution:

The first graph is V shaped graph centered at origin, (0,0) and the graph is not shifted or translated. So, the equation of first graph is,

y=|x|

The second graph is V shaped graph centered at (1,0) and y=|x| is horizontally shifted right by 1 unit. So, the equation of second graph is,

y=|x-1|

The third graph is also V shaped but, it is inverted V graph centered at (0,1) . As, the graph is inverted V So, the graph must have y=-|x|. But, as the center is (0,1). So, the graph is shifted up by 1 unit. So, the equation of the third graph is,

y=-|x|+1

The fourth graph is shifted V graph, whose center is (1,-1). So, y=|x| is shifted right by 1 unit and down by 1 unit. So, the equation of fourth graph is,

y=|x-1|-1

The center of fifth graph is (-1,0), So, y=|x| graph is shifted left by 1 unit. So, the equation of fifth graph is

y=|x+1|

5 0

3 years ago

Read 2 more answers

a boat costs $16600 and decreased in value by 14% per year. How much will the boat be worth after 11 years?

mr_godi [17]

The worth of boat after 11 years is $ 3159.3013

<em><u>Solution:</u></em>

Given that,

Boat costs $16600 and decreased in value by 14% per year

To find: Worth of boat after 11 years

<em><u>The formula for decreasing function is given as:</u></em>

$y = a(1-r)^t$

Where,

y is the value after "t" years

t is the number of years

a is the initial value

r is decreasing rate in decimal

<em><u>From given,</u></em>

a = 16600

t = 11

$r = 14 \% = \frac{14}{100} = 0.14$

<em><u>Substituting the values we get,</u></em>

$y = 16600(1-0.14)^{11}\\\\y = 16600(0.86)^{11}\\\\y = 16600 \times 0.1903\\\\y = 3159.3013$

Thus the worth of boat after 11 years is $ 3159.3013

4 0

3 years ago