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Aleksandr [31]
3 years ago
10

Find the value of 8^2/3

Mathematics
2 answers:
ivann1987 [24]3 years ago
7 0

Answer:

4

Step-by-step explanation:

Calculator

mariarad [96]3 years ago
4 0

Answer:

<h2>4</h2>

Step-by-step explanation:

<h2>8^2/3</h2><h2>2^3(2/3)</h2><h2>2^6/3</h2><h2>2^2</h2><h2>2×2 = 4</h2>
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Lumar needed to pay $5 to enter an arcade. She needs to pay $0.30 for each game she plays.
Marrrta [24]

Answer:

x=game played

y=total amount spent

a) 5+0.3x=y

b) (44-5)/0.3=y

39/0.3=y

130=y

c) 30/0.3=100

6 0
3 years ago
In playing Monopoly, rolling doubles three times in a row sends you to jail. What is the probability of rolling three consecutiv
Alex777 [14]
To find the probability you must multiply the probability of rolling doubles by itself 3 times.

to roll a double  of any number you have to have the same number on a six-sided die that can be represented as 1/6 times 1/6 = 1/36 chance per roll

\frac{1}{36} * \frac{1}{36} * \frac{1}{36} = \frac{1}{46656}

there is a 1 in 46656 chance in rolling doubles 3 times in a row.
8 0
4 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Csf%5Clarge%5Cgreen%7B%5Cunderbrace%7B%5Cred%7BQuestion%2A%7D%7D%7D%3A" id="TexFormula1" tit
Kaylis [27]
<h3><u>Part A</u></h3>

\sf \rightarrow  y^2 - 3y - 4y +12

factor the common term 'y' and '-4'

\sf \rightarrow  y(y - 3) - 4(y - 3)

collect into groups

\sf \rightarrow  (y- 4)(y - 3)

<h3><u>Part B</u></h3>

\sf \rightarrow  9x^2 - 12x + 4

middle term split

\sf \rightarrow  9x^2 - 6x - 6x+ 4

factor the common term '3x' and '-2'

\sf \rightarrow  3x(3x - 2) -2( 3x-2)

collect into groups

\sf \rightarrow  (3x -2)( 3x-2)

simplify the following

\sf \rightarrow  (3x -2)^2

5 0
3 years ago
Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra
katen-ka-za [31]

You made a mistake with the probability p_{j}, which should be p_{i} in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer j, is given by:

(1) \frac{p_{j} }{1-d_{i}p_{i}  } , if j \neq i, and

(2) \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i}  } , if j = i.

so we can say:

A is the event that you search drawer i and find nothing,

B is the event that you search drawer i and find the paper,

C_{k}  is the event that the paper is in drawer k, k = 1, ..., n.

this gives us:

P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}

P(A) = 1 - P(B) = 1 - d_{i} p_{i}

Solution to Part (1):

if j \neq i, then P(A \cap C_{j} ) = P(C_{j} ),

this means that

P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)}  = \frac{P(C_{j} )}{P(A)}  = \frac{p_{j} }{1-d_{i}p_{i}  }

as needed so part one is solved.

Solution to Part(2):

so we have now that if j = i, we get that:

P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}

remember that:

P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}

this implies that:

P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )

so we just need to combine the above relations to get:

P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i}  }

as needed so part two is solved.

8 0
4 years ago
Determine the missing value, x, of the rectangular pyramid below, if the volume is 280 <br> cm3
ExtremeBDS [4]

Answer:

10 cm

Step-by-step explanation:

6 0
3 years ago
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