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AlexFokin [52]
3 years ago
15

How do you write 38% as a fraction, mixed number, or whole number in simplest form? Sorry, incorrect... The correct answer is: E

xplanation review You answered:
Mathematics
1 answer:
MArishka [77]3 years ago
5 0
38/100= 19/50 is the answer to the question
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Is this question a statistical question or a non statistical question: If I
Ber [7]

Answer: Non- Statistical

Step-by-step explanation:

It just isn't

3 0
3 years ago
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G.MG.3 Beth is going to enclose a rectangular area in back of her house. The house wall will form one of the four sides of the f
mina [271]

The maximum area of the rectangular path is when the length is 24 feet and the width is 12 feet.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the length and y represent the width. Hence:

There is 48 feet of fencing:

x + 2y = 48

x  = 48 - 2y   (1)

The area (A) is:

A = xy = y(48 - 2y)

A = 48y - 2y²

The maximum area is at A' = 0, hence:

48 - 4y = 0

y = 12

x = 48 - 2(12) = 24

The maximum area of the rectangular path is when the length is 24 feet and the width is 12 feet.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

6 0
2 years ago
Which pairs of angles in the figure below are vertical angles?
pentagon [3]
A and D
dfngjjdsfhkdfgjhfrreahugjfjdkahghrue Sorry Had to do that for 20 character minumum

7 0
3 years ago
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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
3p . (2+k) = 6p + 3pk
pashok25 [27]
3p* (2+k) = 6p + 3pk

Subtract 6 p + 3 pk from the both sides:

3p(2+k) -(6p+3pk) = 6p+3pk-(6p+3pk)

= 3pk(2+k) - 6p-3pk = 0

Factor 3p(2+k) - 6p - 3k

Factor out common term 3p

= 3p(k-k+2-2)

refine:

0 = 0

true for all p

hope this helps!
3 0
3 years ago
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