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mariarad [96]
2 years ago
5

Two spheres of the same density have a ratio of 4 to 9 in surface area. If the small sphere weighs 10 kg, what does the large sp

here weigh?
Mathematics
2 answers:
Murrr4er [49]2 years ago
8 0
<span>w= 33.75, the weight of the larger sphere
</span>
sertanlavr [38]2 years ago
5 0
We know the surface areas:

Sphere Surface Area     =    <span> 4 • <span>π <span>• r²<span>
radius^2 = Surface Area / 4* PI
</span></span></span></span><span>Let's say area of smaller sphere is 4 PI
radius^2 = 1
radius = 1
9/4 = 2.25
Area of the larger sphere = 4 * PI * 2.25
</span><span>radius^2 = 2.25
radius = 1.5

</span><span>Volume of a sphere = 4/3 • <span>π • r³</span></span>
<span>Volume of larger sphere = 4/3 • <span>π • 1.5^3
</span></span><span>Volume of larger sphere = 3.375 times volume of smaller sphere

</span><span>Mass of larger sphere = 3.375 * 10 kg
The larger sphere weighs 33.75 kg

</span>

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Clare made $160 babysitting last summer. She put the money in a savings account that pays 3% interest per year. If clare doesn't
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We can see that initial value is $160. Upon substituting our given values in above formula, we will get:

y=160(1+0.03)^x

y=160(1.03)^x

To find amount of money in Clare's account after 30 years, we need to substitute x=30 in our equation.

y=160(1.03)^{30}

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2 years ago
RANDOMLY CHOOSING TWO POTENTIAL CLIENTS
gregori [183]
Let <span>Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. </span>

<span>In part IV we are asked:

</span><span>What is the sample space of the pairs of potential clients that could be chosen?
</span><span>
Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:

{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)

          , </span>(C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
<span>                   
</span>                      , (G, M), (G, E), (G, D), (G, A), (G, S)
<span>             
                                    ,</span>(M, E), (M, D), (M, A), (M, S)    
<span> 
                                               , </span>(E, D),  (E, A),  (E, S) <span>   
                                                          
                                                           , </span>(D, A), (D, S)
              
                                                                       , (A, S).}

We can check that the number of the elements of the sample space, n(S) is 

1+2+3+4+5+6+7=28.


This gives us the answer to the first question: <span>How many pairs of potential clients can be randomly chosen from the pool of eight candidates? 

(Answer: 28.)


II) </span><span>What is the probability of any particular pair being chosen?
</span>
The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.

III) <span>What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally? 

The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.


Answers:

I) 28
II) 1/28</span>≈0.0357
III) 1/14≈0.0714
IV)


{(J, C),  (J, G), (J, M),  (J, E),   (J, D),  (J, A),   (J, S)
          , (C, G), (C, M), (C, E),   (C, D), (C, A),  (C, S)
                      , (G, M), (G, E),  (G, D), (G, A),  (G, S)
                                   ,(M, E),  (M, D), (M, A),  (M, S)    
                                               , (E, D),  (E, A),  (E, S)    
                                                            , (D, A), (D, S)
                                                                        , (A, S).}
6 0
2 years ago
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