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wolverine [178]
3 years ago
13

Smaller fraction between 6/13 9/13

Mathematics
2 answers:
My name is Ann [436]3 years ago
7 0

3/13 would be the answer

miss Akunina [59]3 years ago
3 0

\dfrac{6}{13} represents cutting something into 13 pieces and having 6 of those pieces.

\dfrac{9}{13} represents cutting something into 13 pieces and having 9 of those pieces.

Which of those is less?

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PLEASE HELP ME ASAP!!!
Simora [160]

Answer:

C   974,000

Step-by-step explanation:

7.4 × 10^4 + 9.0 × 10^5 =

= 0.74 × 10^5 + 9.0 × 10^5

= 9.74 × 10^5

= 974,000

6 0
2 years ago
A rectangular garden has a walkway around it. The area of
Daniel [21]

Answer:

walkway

around the garden as the sum of two terms

Step-by-step explanation:

A rectangular garden has a walkway around it. The area of

The garden is 6(6.5x+3.5). The combined area of the garden

And the walkway is 6.5(8x+5). Find the area of the walkway

around the garden as the sum of two

5 0
3 years ago
The sum of 3 consecutive even numbers is 132<br><br>pleaase help!!!
agasfer [191]

Answer:

42,44,46

Step-by-step explanation:

we first divide the 132 by the 3 which gives 44

since the numbers involved are even numbers and we know even numbers are numbers divisible by 2 we subtract 2 from the 44 and also add 2 to the 44 the get the rest of the two numbers which is from the above explanation

(44-2),44,(44+2)

42,44,46

we can check whether the above numbers are correct by adding to see whether we get 132

42+44+46=132 which are consecutive even numbers

5 0
3 years ago
Read 2 more answers
26x-4=100 I need the answer to this please
arsen [322]
X=4

26x - 4 = 100
+4 +4
26x = 104
x = 104/26
x=4
5 0
4 years ago
Read 2 more answers
The probability that Chloe acts hungry at 7pm given that she has already eaten dinner is 0.5. The probability that Chloe acts hu
kolezko [41]

Answer:

a) P(  Y^{C} | X ) = 0.180

b) P(Y | X^{C}  ) = 0.998

Step-by-step explanation:

Let

P(X) - Probability that he acts hungry

P(Y) - Probability that he had ate dinner,

Given,

P(X | Y) = 0.5

P(X | Y^{C}  ) = 0.99

P(Y) = 0.9

a.)

P(  Y^{C} | X ) =  \frac{P( X | Y^{C} ). P(Y^{C} )   }{P( X | Y^{C} ). P(Y^{C} ) + P( X | Y ) . P (Y) }

                  = \frac{(0.99)(0.1)}{(0.99)(0.1) + (0.5)(0.9)} = \frac{0.099}{0.549} = 0.180

⇒P(  Y^{C} | X ) = 0.180

b.)

P(Y | X^{C}  ) = \frac{(1 - P(X | Y ) ) . P(Y)}{P( X^{C} )  } =  \frac{(1 - P(X | Y ) ) . P(Y)}{ 1 - P( X )  }

                 = \frac{(1 - 0.5)(0.9)}{1 - [(0.99)(0.1) + (0.5)(0.9) ]} = \frac{(0.5)(0.9)}{1 - [(0.099) + (0.45) ]} = \frac{0.45}{1 - [0.540]} = \frac{0.45}{0.451} = 0.998

⇒P(Y | X^{C}  ) = 0.998

3 0
3 years ago
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