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Verizon [17]
4 years ago
5

Help on this problem?? please?

Mathematics
1 answer:
Reptile [31]4 years ago
5 0

MrBillDoesMath!


Answer to #4:  81/256 * s^8 * t^ 12


Comments:

(7x^3) ^ (1/2)   =  7 ^ (1/2)  *  x^(3/2)   where  ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.

---------------------

(1)             (27s^7t^11)^ (4/3)  

               = 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)


As 27 = 3^3, 27 ^(4/3) = 3^4 = 81

               

(2)  (-64st^2)^ (4/3)  =     (-64)^(4/3) * (s^4/3) * t(^8/3)


As 64 = (-4)^3,  (-64)^(4/3) = (-4)^4 = +256                                        


So (1)/(2) =

81 * s^(28/3)* t^(44/3)

-------------------------------     =

256 s^(4/3) * t^((8/3)


81/256 *  s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =


81/256 * s^(24/3) * t (36/3) =

81/256 * s^8        * t^ 12



MrB

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p
Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is  \frac{23}{30} or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

P(A \cup B)=P(A)+P(B)-P(A \cap B)---------------(1)

where

P(A) = Probability of choosing  a student playing basket ball

P(B) =  Probability of choosing  a student playing base ball

P(A \cap B) =  Probability of choosing  a student playing both the games

<u>Finding  P(A)</u>

P(A) = \frac{\text { Number of students playing basket ball }}{\text{Total number of students}}

P(A) = \frac{19}{30}--------------------------(2)

<u>Finding  P(B)</u>

P(B) = \frac{\text { Number of students playing baseball }}{\text{Total number of students}}

P(B) = \frac{12}{30}---------------------------(3)

<u>Finding P(A \cap B)</u>

P(A) = \frac{\text { Number of students playing both games }}{\text{Total number of students}}

P(A) = \frac{8}{30}-----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}

P(A \cup B)= \frac{31}{30} -\frac{8}{30}

P(A \cup B)= \frac{23}{30}

7 0
3 years ago
What is the fraction 18/24 reduced to its lowest terms?
WINSTONCH [101]
<em>First, find the greatest common factor (GCF) of the numerator and denominator.</em>

<u>Factors of 18</u>: 1, 2, 3, 6, 9, 18
<u>Factors of 24</u>: 1, 2, 3, 4, 6, 8, 12, 24
<u>Common Factors</u>: 1, 2, 3, 6
<u>GCF</u>: 6

<em>Now, divide the numerator by 6 and the denominator by 6.</em>

18 ÷ 6 = 3
24 ÷ 6 = 4

<em>Set these as your new numerator and denominator.</em>

\frac{3}{4}

The answer is (b).

8 0
3 years ago
Read 2 more answers
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