Class A has 9 pupils and class B has 12 pupils.

Help please!!!

a_sh-v [17]

Answer:

yes thats good

Step-by-step explanation:

4 0

3 years ago

Which value can be added to the set below without changing its mean?

shusha [124]

Answer:

15..

Step-by-step explanation:

4, 10, 14, 18, 22, 22

Mean = (4 + 10 + 14 + 18 + 22 + 22)/6

= 90/6

= 15.

Number to be added, 'y'

For the mean to be the same

(90 + y) / 7 = 15

90 + y = 15 × 7

90 + y = 105

y = 105 - 90

= 15

7 0

3 years ago

Rewrite the function by completing the square<br><br> h(x)=4x^2+4x+1

lisov135 [29]

Answer:

$h(x) = 4(x+\frac{1}{2})^{2}$

Step-by-step explanation:

$h(x)=4x^2+4x+1\\h(x)=4(x^2+x+\frac{1}{4}) \\h(x)=4(x^2+x+\frac{1}{4})\\h(x) = 4(x+\frac{1}{2})^{2}$

8 0

3 years ago

One plant from the garden is randomly selected. There is a 30% chance that the plant was purchased this year, and there is a 6%

Juliette [100K]

Answer and workings in the attachment below. Also, do me one big favour. Let me know if it's the answer your teacher or text book ends up giving you. That would give me some satisfaction.

Answer: 80%

7 0

3 years ago

Read 2 more answers

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h

mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere $=\frac{2}{3}\pi r^3$

Volume of a Cylinder $=\pi r^2 h$

Therefore:

The Volume of the Solid formed

$=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Area of the Hemisphere = $2\pi r^2$

Curved Surface Area of the Cylinder = $2\pi rh$

Total Surface Area=

$2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh$

Cost of the Hemispherical Ends = 2 X Cost of the surface area of the sides.

Therefore total Cost, C

$=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh$

Recall: $h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Therefore:

$C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}$