What is the area of this figure? Enter your answer in the box. mm²

When Ximena commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 38 minutes and a

irga5000 [103]

Answer:

The interval that represents the middle 68% of her commute times is between 33.5 and 42.5 minutes.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 38 minutes, standard deviation of 4.5 minutes.

Determine the interval that represents the middle 68% of her commute times.

Within 1 standard deviation of the mean. So

38 - 4.5 = 33.5 minutes

38 + 4.5 = 42.5 minutes.

The interval that represents the middle 68% of her commute times is between 33.5 and 42.5 minutes.

6 0

3 years ago

Solve the formula V=pir^2h for r PLEAASSSEEE HELP

otez555 [7]

Answer:

B

Step-by-step explanation:

So we have the formula:

$V=\pi r^2h$

And we want to solve it for r.

So, let's first divide both sides by π and h. This will cancel out the right side:

$r^2=\frac{V}{\pi h}$

Now, take the square root of both sides:

$r=\sqrt{\frac{V}{\pi h}}$

And we're done!

Our answer is B.

I hope this helps!

7 0

3 years ago

Read 2 more answers

What is the greatest common factor for 200,205

bixtya [17]

$200|2\\100|2\\.\ 50|2\\.\ 25|5\\.\ \ 5|5\\.\ \ 1|\\\\200=2\times2\times2\times\fbox5\times5\\-----------------\\205|5\\.\ 41|41\\.\ \ 1|\\\\205=\fbox5\times41\\-------------------\\\\GCF(200;\ 205)=\fbox5$

6 0

3 years ago

Consider the following pair of equations:

Aloiza [94]

Y = 3x + 3 this is equation 1
y = x -1 this is equation 2
since equation 1 and 2 already defined y in terms of x, we're just gonna substitute y into the other equation. as 2 is shorter I'm going with that today
x - 1 = 3x +3
-1 - 3 = 3x - x
-4 = 2x
x = -2
now we've got x just sub it in equation 2
y = -2 -1
y = -3
so the answer is (-2,-3)

5 0

3 years ago

*Asymptotes* g(x) =2x+1/x-3 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