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faust18 [17]
2 years ago
13

Find the hight of cylinder with a volume of 100cm3 and a radius of 2 cm??​

Mathematics
2 answers:
Novosadov [1.4K]2 years ago
7 0

Answer:

7.95775

Step-by-step explanation:

volume = pi*r^2*h

Using the formula

V=πr2h

Solving for h

h=V/πr2

=100/π·2^2≈7.95775

Nastasia [14]2 years ago
5 0

Answer in attachment......

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Which strategy would not correctly solve this story problem? Jason bought a bagel and a cup of coffee each Saturday. Each bagel
snow_tiger [21]
This answer is not correct

<span>C. Use objects to model the problem. Take 8 pieces of play money. Take away 1 piece for the coffee and 2 pieces for the bagel.</span>
8 0
3 years ago
224 is what % of 640?
lakkis [162]
There is a percent formula for all of these types of percent problems.

It is : is/of = %/100

224 is what % of 640
is = 224
of = 640
% = x
now just sub
224/640 = x/100
cross multiply because this is a proportion
(640)(x) = (224)(100)
640x = 22400
x = 22400/640
x = 35......so 224 is 35% of 640


5 0
3 years ago
Read 2 more answers
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andrew-mc [135]

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6 0
2 years ago
Follow the process of completing the square to solve 2x2 + 8x - 12 = 0. After adding B2 to both sides of the equation in step 4,
valentina_108 [34]
 2x2 + 8x - 12 = 0
4 + 8x - 12 = 0
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4 0
3 years ago
Urn 1 contains 3 blue tokens and 2 red tokens; urn 2 contains 2 blue tokens and 4 red tokens. All tokens are indistinguishable.
Dmitry_Shevchenko [17]

Answer:

RR = 0.4

RB = 0.3

BB = 0.22

BR = 0.30

Step-by-step explanation:

P( Urn 1 ) = 2/6 = 1/3

P( Urn 2 ) = 1 - 1/3 = 2/3

Urn 1 contains : 3 blue and 2 red

P( blue | urn 1 ) = 3/5 ( with replacement ) , P( blue | urn 1 ) = 3/4 ( without replacement )

P( red | urn 1 ) = 2 / 5 ( with replacement ) , P(red | urn 1 ) = 1/2 ( without replacement )

Urn 2 contains : 2 blue and 4 red

P ( blue | urn 2 ) = 1/3 ( with replacement ) , P( blue | urn 2 ) = 2/5 ( without replacement )

P ( red | urn 2 ) = 2/3 ( with replacement) , P( red | urn 2 ) = 4/5 ( without replacement )

Determine

<u>i) Possible outcomes when two tokens are drawn from either Urn without replacement </u>

RR = [[ ( 2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] = 0.4

RB = [[ (2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.3

BB = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.22

BR = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] ≈ 0.30

<u />

8 0
3 years ago
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