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3 years ago

6

Sectionalizing formula of 2

ign="absmiddle" class="latex-formula">

1 answer:

kozerog [31]3 years ago

3 0

Answer:

$\sqrt{3 \times 2 {?}^{2} } = \sqrt{12} = 3.46$

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If 6 men finish a piece of work in 8 days, how long will it rake 4 men to

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Answer:

6 men finish a piece of work in 8 days

1 men finish a piece of work in 8*6 days

4men will finish a piece of work in 48/4

days

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which is equal to 12 days

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What is 18 times 40 to the 9th power

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I believe thugs is it
4.718592e+15

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Please someone help me on this <33

dangina [55]

Answer:

B

Step-by-step explanation:

The initial value for a function is the value of y when x=0. For function A, we can see from the table that when x = 0, we have y = 4. For function B, we plug in 0 in our equation and solve: y = 3(0) + 5, or y = 5. So, because function B has a larger value of y when x = 0, function B has the greater initial value because the initial value for Function A is 4 and the initial value for function B is 5.

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3 years ago

4. The following intervals were plausible population proportions for a given sample. Find the margin of error in each case.

Vedmedyk [2.9K]

Answer: a. 0.15 b. 0.03 c. 0.055 d. 0.05

Step-by-step explanation:

For plausible intervals of population proportions , the margin of error is basically half of the difference between upper limit and lower limit.

a. Interval = From 0.35 to 0.65

Upper limit = 0.65

Lower limit = 0.35

Difference = Upper limit- Lower limit

= 0.65-0.35=0.30

Margin of error = (Difference)÷2 =(0.30)÷2=0.15

b. Interval = From 0.72 to 0.78

Upper limit = 0.78

Lower limit = 0.72

Difference = 0.78-0.72 = 0.06

Margin of error =(0.06)÷2=0.03

c. Interval = From 0.84 to 0.95

Upper limit = 0.95

Lower limit = 0.84

Difference =0.95-0.84= 0.11

Margin of error = (0.11)÷2 = 0.055

d.Interval = From 0.47 to 0.57

Upper limit = 0.57

Lower limit = 0.47

Difference = 0.57-0.47=0.10

Margin of error = (0.10)÷2=0.05

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3 years ago

A jumbo crayon is composed of a cylinder with a conical tip. The cylinder is 12 cm tall with a radius of 1.5 cm, and the cone ha

s2008m [1.1K]

Answer:

Part 1) The lateral area of the cone is $LA=2\pi\ cm^{2}$

Part 2) The lateral surface area of the cylinder is $LA=36\pi\ cm^{2}$

Part 3) The surface area of the crayon is $SA=41.50\pi\ cm^{2}$

Step-by-step explanation:

Part 1) Find the lateral area of the cone

The lateral area of the cone is equal to

$LA=\pi rl$

we have

$r=1\ cm$

$l=2\ cm$

substitute

$LA=\pi (1)(2)$

$LA=2\pi\ cm^{2}$

Part 2) Find the lateral surface area of the cylinder

The lateral area of the cylinder is equal to

$LA=2\pi rh$

we have

$r=1.5\ cm$

$h=12\ cm$

substitute

$LA=2\pi (1.5)(12)$

$LA=36\pi\ cm^{2}$

Part 3) Find the surface area of the crayon

The surface area of the crayon is equal to the lateral area of the cone, plus the lateral area of the cylinder, plus the top area of the cylinder plus the bottom base of the crayon

<em>Find the area of the bottom base of the crayon</em>

$A=\pi[r2^{2}-r1^{2}]$

where

r2 is the radius of the cylinder

r1 is the radius of the cone

substitute

$A=\pi[1.5^{2}-1^{2}]$

$A=1.25\pi\ cm^{2}$

<em>Find the area of the top base of the cylinder</em>

$A=\pi(1.5)^{2}=2.25\pi\ cm^{2}$

<em>Find the surface area</em>

$SA=2\pi+36\pi+2.25\pi+1.25\pi=41.50\pi\ cm^{2}$

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