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Black_prince [1.1K]
3 years ago
13

Liner equations with one variable!! Please show work, ill give brainliest

Mathematics
1 answer:
matrenka [14]3 years ago
8 0
This is hard to solve
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A board 1.2 m long is cut into two so that the length of the longer piece is 15cm longer than twice the length of the shortest p
MAXImum [283]

Answer:

The length of the shorter piece=0.35 m

Step-by-step explanation:

Let the lengths be as follow;

Shorter piece=x

Longer piece=15 cm longer than twice shorter piece(x)

Since 1 m=100 cm, 15 cm=15/100=0.15 m

Longer piece= (2×x)+0.15=2x+0.15

Total length=1.2 m

Total length=shorter piece+longer piece

Replacing;

1.2=x+2x+0.15

3x=1.2-0.15

3x=1.05

x=(1.05/3)=0.35

The length of the shorter piece=x=0.35

4 0
3 years ago
Based on this equation, estimate the price of a car that had been driven 56 thousand kilometers.
OLEGan [10]

Answer:

50 dollars

Step-by-step explanation:

7 0
2 years ago
There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl
insens350 [35]

We're told that

P(A\cap B)=0.15

P(A\cup B)^C=0.06\implies P(A\cup B)=0.94

P(B\mid A)=P(B^C\mid A)=0.5

where the last fact is due to the law of total probability:

P(A)=P(A\cap B)+P(A\cap B^C)

\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)

\implies 1=P(B\mid A)+P(B^C\mid A)

so that B\mid A and B^C\mid A are complementary.

By definition of conditional probability, we have

P(B\mid A)=P(B^C\mid A)

\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}

\implies P(A\cap B)=P(A\cap B^C)

We make use of the addition rule and complementary probabilities to rewrite this as

P(A\cap B)=P(A\cap B^C)

\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)

\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)

\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]

\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]

\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C

\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)

By the law of total probability,

P(B)=P(A\cap B)+P(A^C\cap B)

\implies P(A^C\cap B)=P(B)-P(A\cap B)

and substituting this into (*) gives

2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]

\implies P(B)=P(A\cup B)-P(A\cap B)

\implies P(B)=0.94-0.15=\boxed{0.79}

8 0
3 years ago
During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim
Maru [420]

Answer:

1) Juans claim is incorrect. The correct experimental probablilty is 2/9

Step-by-step explanation:

8 0
3 years ago
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PLZ HELP 3 MINUTES LEFT WIL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jobisdone [24]

Answer:

the first box

Step-by-step explanation:

7 0
2 years ago
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