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2 years ago

5

Q.2. A cricketer scores the following runs in eight innings :

1 answer:

dangina [55]2 years ago

7 0

Answer: Mean Score = 50 Range = 31

Step-by-step explanation:

58+76+40+35+46+45+0+100=400

400/8=50

Range = subtract smallest number from largest number

54 - 23 = 31

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BEST ANSWER GETS BRAINLIEST!!!!!!!!

Ad libitum [116K]

Answer:

(x + 6, y + 0), 180° rotation, reflection over the x‐axis

Step-by-step explanation:

The answer can be found out simply , a trapezoid has its horizontal sides usually parallel meanwhile the vertical sides are not parallel.

The horizontal parallel sides are on the x-axis.

Reflection over y- axis would leave the trapezoid in a vertical position such that the trapezoid ABCD won't be carried on the transformed trapezoid as shown in figure.

So option 1 and 2 are removed.

Now, a 90 degree rotation would leave the trapezoid in a vertical position again so its not suitable again.

In,The final option (x + 6, y + 0), 180° rotation, reflection over the x‐axis, x+6 would allow the parallel sides to increase in value hence the trapezoid would increase in size,

180 degree rotation would leave the trapezoid in an opposite position and reflection over x-axis would bring it below the Original trapezoid. Hence, transformed trapezoid A`B`C`D` would carry original trapezoid ABCD onto itself

8 0

2 years ago

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0

3 years ago

Help me with this question, please?

Virty [35]

Answer:vertices

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

R+q-r; use q=6,and r=2

Anastasy [175]

Answer:

so you would replace all r with 2 which is 2+q-2

then replace the q with 6 so it would be: r+6-r

combined both of these to get

2+6-2

which would equal to 6

add the 2 and the 6 to get 8

then subtract 2 to get 6

3 0

3 years ago

Read 2 more answers

Please help asap. BRAINLIEST--

son4ous [18]

D. Rectangle
It will be smaller than the base

8 0

2 years ago