Please help with three questions!! Worth 98 points!!

I'm assuming a quarter-circle is exactly 1/4 of a circle. Thus if you have 4 congruent quarter-circles, that should mean they make a complete circle.

If that is the case, then we can find the area of the full circle using pi*r^2.

So the area of the circle is 5^2*pi or 25pi.

To find the area of the shaded region, we subtract the area of the circle from the area of the square.

The area of the square is 10^2 or 100.

So the area of the shaded region is 100 - 25pi.

My calculator says that equals roughly 21.46

6 0

3 years ago

Consider the following information related to a set of quantitative sample data:

svlad2 [7]

Answer:

(a) There are outliers

(b) $x$ and $x >62$

Step-by-step explanation:

Given

$\sigma = 14.92$

$\bar x = 22.0$

$q_0 = -24$

$q_1 = 14.5$

$q_2 = 24.5$

$q_3 = 33.5$

$q_4 = 64$

Solving (a): Check for outliers

This is calculated using:

$Lower = Q_1 - (1.5 * IQR)$ --- lower bound of outlier

$Upper = Q_3 +(1.5 * IQR)$ --- upper bound of outlier

Where

$IQR = Q_3 - Q_1$

So, we have:

$IQR = 33.5 - 14.5$

$IQR = 19$

The lower bound of outlier becomes

$Lower = Q_1 - (1.5 * IQR)$

$Lower = 14.5 - (1.5 * 19)$

$Lower = 14.5 - 28.5$

$Lower = -14$

The upper bound of outlier becomes

$Upper = Q_3 +(1.5 * IQR)$

$Upper = 33.5 + 1.5 * 19$

$Upper = 33.5 + 28.5$

$Upper = 62$

So, we have:

$-14 \le x \le 62$ --- the range without outlier

Given that:

$q_0 = -24$ --- This represents the lowest data

$q_4 = 64$ --- This represents the highest data

-24 and 64 are out of range of $-14 \le x \le 62$ .

Hence, there are outliers

Solving (b): The outliers

The outliers are data less than the lower bound (i.e. less than -14) or greater than the upper bound (i.e. 62)

So, the outliers are:

$x$ and $x >62$

4 0

2 years ago

How to find average value of a function over a given interval?

Strike441 [17]

f(x)=8x−6f(x)=8x-6 , [0,3][0,3]

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.(−∞,∞)(-∞,∞){x|x∈R}{x|x∈ℝ}f(x)f(x) is continuous on [0,3][0,3].f(x)f(x) is continuousThe average value of function ff over the interval [a,b][a,b] is defined as A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dx.A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dxSubstitute the actual values into the formula for the average value of a function.A(x)=13−0(∫308x−6dx)A(x)=13-0(∫038x-6dx)Since integration is linear, the integral of 8x−68x-6 with respect to xx is ∫308xdx+∫30−6dx∫038xdx+∫03-6dx.A(x)=13−0(∫308xdx+∫30−6dx)A(x)=13-0(∫038xdx+∫03-6dx)Since 88 is constant with respect to xx, the integral of 8x8x with respect to xx is 8∫30xdx8∫03xdx.A(x)=13−0(8∫30xdx+∫30−6dx)A(x)=13-0(8∫03xdx+∫03-6dx)By the Power Rule, the integral of xx with respect to xx is 12x212x2.A(x)=13−0(8(12x2]30)+∫30−6dx)

3 0

3 years ago