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3 years ago

How to write a fraction or mixed number as a decimal pre algebra?

1 answer:

4 0

If you had the number 4 1/4 you would divide the numerator (1) by the denominator (4) and then take the decimal number (0.25) you get and add the decimal to the whole number on your mixed number (4+0.25) and that would give you the decimal.

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The length of one leg of a right triangle is 8 centimeters shorter than the

ra1l [238]

${34}^{2} + {b}^{2} = {42}^{2} \\ 1156 + {b}^{2} = 1764 \\ 1156 - 1156 + {b}^{2} = 1764 - 1156 \\ {b}^{2} = 608 \\ \sqrt{ {b}^{2} } = \sqrt{608} \\ b = 24.7$
I used the Pythagorean formula, which is a² + b² = c².
a² and b² are the two side legs while c² is the hypotenuse.
One of the side is 42 - 8 = 34 cm.

3 0

3 years ago

Answer ASAP PLS -

Talja [164]

TQV: 50
good luck :’)

3 0

3 years ago

A basketball rim 10 ft high casts a shadow 15 ft long. at the same time, a nearby building casts a shadow that is 54 ft long. ho

Tatiana [17]

15/10= 3/2
54/x=3/2
3× what=54
18
18×2= 36 ft the building is 36 ft

7 0

3 years ago

Somebody please help me

WITCHER [35]

Step-by-step explanation:

x - 1 = - 3x - 14

Bringing like terms on one side

x + 3x = -14 + 1

4x = - 13

x = - 13/4

5 0

3 years ago

. Imagine a game of 3 players where exactly one player wins in the end and all players have equal chances of being the winner. T

lbvjy [14]

Answer:

5/9

Step-by-step explanation:

Number of players = 3

number of times game is repeated = 4

P( any person wins a game ) = 1/3

P ( any person does not win a game ) = 1 - 1/3 = 2/3

P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81

Note : each player has equal chance of winning

Find the probability that there is at least one person who wins no games

lets represent the probability of each player not wining a game with alphabet A

A1 = player 1 wins no game

A2 = player 2 wins no game

A3 = players 3 wins no game

Applying the inclusion-exclusion formula

P( A1 ∪ A2 U A3 ) = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1 ∩ A3 ) + P( A1 ∩ A2 ∩ A3 )

where

P( A1 ∩ A2 ) = P( A1 wins all games )

P ( A1 wins all games in 4 attempts ) = ( 1/3 )^4 = 1/81

P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0

Hence

P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9

6 0

3 years ago