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UkoKoshka [18]
3 years ago
12

Can you do question c for me?

Mathematics
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer:

28.2

Step-by-step explanation:

91 × 18 = 1638

1638 ÷ 58 = 28.2413793103

rounded to 1 significant figure = 28.2 as 4 rounds down

hope this helps

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Solve for x. 4.25x = 21.
Alex787 [66]

Answer:21 divided by 4.25 = 4.941176471

Step-by-step explanation:

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3 years ago
When the product of 6 and the square of a number is increased by 5 times the number, the result is 4. which equation represents
AleksandrR [38]

Answer:

The answer would be 6x^2 + 5x - 4 = 0

4 0
2 years ago
B) Simplify 9y - 6y + y
julsineya [31]

Answer:

The answer is 4y

Step-by-step explanation:

Simplify the expression.

Hoped this helped!

Could I perhaps get brainly?

8 0
2 years ago
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What is the approximate solution of the linear system represented by the graph below
VashaNatasha [74]
ANSWER

The approximate solution is

(4, - 1)

EXPLANATION

The point of intersection of the two lines is the solution of the linear systems represented by the graph.


From the graph we can determine that each box is 2 units.


From the graph, when we start at O and count two boxes to the right and half box down, we obtain the point of intersection of the two lines to be
(4, - 1)


Note that since each box is two units, half way between the boxes gives one unit.

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

<em>Note : each player has equal chance of winning </em>

<u>Find the probability that there is at least one person who wins no games </u>

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

<em>P( A1 ∪ A2 U A3 )</em><em> = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1            ∩ A3 )  + P( A1 ∩ A2 ∩ A3 ) </em>

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
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