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2 years ago

5

Marie wants to know the most popular car for parents. She plans to survey the people entering the mall from 10:00 a.m. to 11:00

p.m. What can Marie do to make sure her survey is valid?
A. Marie's sample is valid.
B. Marie should also survey people entering the grocery stores.
C. Marie should make sure she surveys both the fathers and mothers.
D. Marie should make sure she surveys the children as well as the parents.

1 answer:

Andreyy892 years ago

3 0

Answer:

C. Marie should make sure she surveys both the fathers and mothers.

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Solve for x in each of the equations or inequalities below, and name the property and/or properties used:

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a) 3/4 x = 9

x = $\dfrac{9\times 4}{3}$

x = 12

b) 10+ 3x = 5x

2 x = 10

x = 5

c) a + x = b

x = b - a

d) c x = d

x = $\dfrac{d}{c}$

e) $\dfrac{1}{2} x - g < m$

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3 years ago

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

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