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1 year ago

I need to send photos.

Mathematics

1 answer:

frez [133]1 year ago

6 0

SOLUTION:

Step 1:

From the given question, and comparing the scores in classes A and B

Step 2:

Question one, to know the class which had the better overall result on the exam;

(I) The scores in class A ranges from 65 to 100, while that of class B ranges from 60 to 90. This explains that class A had better results.

(ii) The median of the scores in class A is 85, while the median of the scores in class B is 75, this is another piece of supporting evidence that class A had better results.

Step 3:

Question two, to know the class which had greater variability in the results;

For a better understanding of this question, I need to explain the concept of variability in statistics.

Variability in statistics refers to the difference being exhibited by data points within a data set, as related to each other or as related to the mean. This can be expressed through the range, variance or standard deviation of a data set.

Step 4:

So we need to find the range of scores in each of the two classes and then compare, the class with the greater range has the greater variability.

The Range is the difference between the lowest and highest score (H - L), where H is the highest score and L is the lowest score.

Step 5:

Applying the formula for range stated in step 4;

For Class A; H = 100 and L = 65

For Class B; H = 90 and L = 60

The range for class A; H - L = 100 - 65 = 35

The range for class B; H - L = 90 - 60 = 30

By comparing the range of class A and that of B, it is clear that Class A had a greater range (variability)

CONCLUSION:

Class A had better overall results in the exam and greater variability in the results.

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A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 56 ft

harkovskaia [24]

Answer:

Dimensions of the window in order to allow maximum light is $x=\dfrac{56}{4+\pi}$ and $y=\dfrac{56}{4+\pi}$

Step-by-step explanation:

Consider following Norman window, assuming ABCD as rectangle and arc AD as semicircle with center at E and radius r. (Refer attachment)

Given that perimeter of window is 56 ft. Therefore perimeter of window is given as,

$Perimerter = AB + BC + CD + arc\:AD$

Calculate arc AD as follows,

Let, x denote radius of semi circle. That is, r=x

Since AD is the diameter of semi circle

So AD = 2 r = 2 x.

Now perimeter of semicircle is equal to circumference of semicircle, so calculate circumference of semicircle as follows

Circumference of circle is $C=2\pi r$ . So half of it will be

$C=\dfrac{2\pi r}{2}$

$C=\dfrac{2\pi x}{2}$

$C=\pi x$

So, $arc\:AD = \pi x$

Calculate AB , BC and AD as follows,

Consider rectangle ABCD,

Since, AD is the diameter of semi circle which is also one of the side of the rectangle.

So AD = 2 r = 2 x.

Since AD is parallel to BC. Therefore, AD=BC=2x. Also length of rectangle be y

$\therefore BC=2x,AB=CD=y$

Substituting the value,

Perimeter = AB + BC + CD + arc AD

$56=y+2x+y+\pi x$

$56=2y+2x+\pi x$

To calculate value of y,subtracting 2x and $\pi x$ on both sides,

$56-2x-\pi x=2y$

Dividing by 2,

$28-x-\dfrac{\pi x}{2}=y$

Now calculate area of window.

Area = Area of rectangle + Area of semicircle

From diagram,

$Area = width\times length + \dfrac{1}{2}\times\pi r^{2}$

$Area = 2x\times y + \dfrac{1}{2}\times\pi x^{2}$

$Area = 2xy + \dfrac{\pi x^{2}}{2}$

Substituting value of y in above equation,

$Area = 2x\left (28-x-\dfrac{\pi x}{2} \right ) + \dfrac{\pi x^{2}}{2}$

Simplifying,

$Area = 56x-2x^{2}-\pi x^{2}+ \dfrac{\pi x^{2}}{2}$

$Area = 56x-2x^{2}- \dfrac{\pi x^{2}}{2}$

In order to find the maximum of area function, differentiate the equation with respect to x and find the critical points.

Applying difference rule of derivative,

$\dfrac{dA}{dx}=\dfrac{d}{dx}\left(56x\right)-\dfrac{d}{dx}\left ( 2x^{2} \right )-\dfrac{d}{dx}\left ( \dfrac{\pi x^{2}}{2} \right )$

Applying constant multiple rule of derivative,

$\dfrac{dA}{dx}=56\dfrac{d}{dx}\left(x\right)-2\dfrac{d}{dx}\left ( x^{2} \right )-\dfrac{\pi}{2}\dfrac{d}{dx}\left (x^{2}\right )$

Applying power rule of derivative,

$\dfrac{dA}{dx}=56\left(1x^{1-1}\right)-2\left(2x^{2-1}\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x^{2-1}\right )$

$\dfrac{dA}{dx}=56\left(1\right)-2\left(2x\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x\right )$

$\dfrac{dA}{dx}=56-4x-\pi x$

Now find the critical number by solving as follows,

$\dfrac{dA}{dx}=0$

$56-\left(4+\pi\right)x =0$

$56=\left(4+\pi \right)x$

$\dfrac{56}{4+\pi} =x$

Since there is only one critical point, directly substitute the value of x into equation of A. If value of A is greater than 0, then the area is maximum at critical point.

$Area = 56\left (\dfrac{56}{4+\pi} \right )-2\left (\dfrac{56}{4+\pi} \right )^{2}- \dfrac{\pi \left (\dfrac{56}{4+\pi} \right )^{2}}{2}$

Calculating the above expression,

$Area = \dfrac{1568}{4+\pi }$

So area is greater than 0.

Now calculate value of y,

$28-\dfrac{56}{4+\pi}-\dfrac{\pi}{2}\left ( \dfrac{56}{4+\pi} \right )=y$

$\dfrac{56}{4+\pi}=y$

Hence dimensions are $x=\dfrac{56}{4+\pi}$ and $y=\dfrac{56}{4+\pi}$

3 0

4 years ago

I need help on this question

tatiyna

It is a right agn have a good day

6 0

3 years ago

15.25 rounded to the nearest tenth?

Airida [17]

Rules for rounding: If a number is less than 5, round down. If it is 5 or bigger, round up. Tenths is the one just to the right of the decimal. Since the hundredths is 5, we round up. The answer is 15.3

:)-Kibeye

8 0

4 years ago

Read 2 more answers

What is the slope of the line?

VladimirAG [237]

Answer:

-1

Step-by-step explanation:

Pick two points on the line

(0,1) and ( 2,-1)

Using the slope formula

m = (y2-y1)/(x2-x1)

= ( -1 -1)/(2-0)

= -2/2

= -1

5 0

3 years ago

What is the solution to the system of equations 8x-10y=-23 and 9x+10y=-16

sammy [17]

8x-10y=-23
9x+10y=-16
(8x+9x)+(10y-10y)=(-23+(-16))
17x+0y=-39
x=-39/17
8(-39/17)-10y=-23
-312/17-10y=-23
-10y=-23+312/17
-10y=-79/17
y=79/170

3 0

4 years ago