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Alexandra [31]
2 years ago
11

Solve to find X. 1/5(x - 2) = 1/10 (x + 6)

Mathematics
2 answers:
Archy [21]2 years ago
5 0

Answer:

Step-by-step explanation:

1/5(x-2) =1/10(x+6)

1/5x-2 = 1/10x+0.6

1/5x+1/10x= 0.6-2

0.2x+0.1x= -1.4

0.3x= -1.4

x= -4.67

x= -4.7

x= -5

LenaWriter [7]2 years ago
3 0

Answer:

x = 10

Step-by-step explanation:

First, distribute:

\frac{1}{5} x - \frac{2}{5} =\frac{1}{10} x+\frac{6}{10}

Next, subtract \frac{1}{5}x from both sides:

-\frac{2}{5} = \frac{1}{10} x -\frac{1}{5} x+\frac{6}{10}

(multiply \frac{1}{5}x by 2 to get common denominator, \frac{2}{10} x )

-\frac{2}{5} =-\frac{1}{10}x +\frac{6}{10}

Next, subtract \frac{6}{10} from both sides:

-\frac{2}{5} -\frac{6}{10} =-\frac{1}{10} x

(multiply \frac{2}{5} by 2 to get common denominator, \frac{4}{10} )

-\frac{10}{10} =-\frac{1}{10} x ⇒ -1=-\frac{1}{10} x

Next, multiply both sides by 10:

-10=-x

FInally, divide both sides by -1:

x = 10

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an artificial lake is in the shape of a rectangle and has an area of 9/20 square mile. the width of the lake is 1/5 the length o
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X=length
y=width
Area (rectangle)=length x width
we suggest this system of  equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2   it does not validate
x₂=3/2    ⇒y=x/5=(3/2)/5=3/10

The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles

to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒  3/10 miles=(3/2) /5 miles

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F(x)= (x+9x2- 9x +15
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Answer:

10x+15

Step-by-step explanation:

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In a particular course, it was determined that only 70% of the students attend class on Fridays. From past data it was noted tha
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Answer: Probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

Step-by-step explanation:

Since we have given that

Probability that students attend class on Fridays = 70% = 0.7

Probability that who went to class on Fridays would pass the course = 95% = 0.95

Probability that who did not go to class on Fridays would passed the course = 10% = 0.10

Let A be the event students passed the course.

Let E be the event that students attend the class on Fridays.

Let F be the event that students who did not attend the class on Fridays.

Here, P(E) = 0.70 and P(F) = 1-0.70 = 0.30

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We need to find the probability that they did not attend on Fridays.

We would use "Bayes theorem":

P(F\mid A)=\dfrac{P(F).P(A\mid F)}{P(E).P(A\mid E)+P(F).P(A\mid F)}\\\\P(F\mid A)=\dfrac{0.30\times 0.10}{0.70\times 0.95+0.30\times 0.10}\\\\P(F\mid A)=\dfrac{0.03}{0.695}=0.043

Hence, probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

8 0
4 years ago
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