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2 years ago

3(2y-4) - 2 ) = 0.4 (1+y)

Mathematics

1 answer:

Lelu [443]2 years ago

3 0

6y-12-2=0.4+0.4y
6y-10-0.4=0.4-0.4+0.4y
6y-6y-9.6=0.4y-6y
(-9.6=-5.6y)/5.6
1.7143=y

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(18x-44 )(8x-10) (13y-38) geometry

xenn [34]

Simplifying

18x + -44 = 13y + -38

Reorder the terms:

-44 + 18x = 13y + -38

Reorder the terms:

-44 + 18x = -38 + 13y

Solving

-44 + 18x = -38 + 13y

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '44' to each side of the equation.

-44 + 44 + 18x = -38 + 44 + 13y

Combine like terms: -44 + 44 = 0

0 + 18x = -38 + 44 + 13y

18x = -38 + 44 + 13y

Combine like terms: -38 + 44 = 6

18x = 6 + 13y

Divide each side by '18'.

x = 0.3333333333 + 0.7222222222y

Simplifying

x = 0.3333333333 + 0.7222222222y

3 0

3 years ago

If the list price of bicycle is Rs 1000 and it is sold for Rs 900,cfind the list price of the article.

fgiga [73]

Answer:

Accroding to the question my guess for the answer would be 10

Step-by-step explanation:

I solved it like this

1000-900

= 100

=1000/100

=10

6 0

2 years ago

Emma works on an ice cream truck. She has lots of ice creams in the cooler. On Friday, she gave out 58 ice creams. Now she has 7

damaskus [11]

Answer:

She had a total of 131 before she sold anything

Step-by-step explanation:

Add 58 and 73

4 0

3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

Represent the following sentence as an algebraic expression, where "a number" is the letter x. You do not need to simplify.

My name is Ann [436]

<h3>Answer: (2x)^3</h3>

Work Shown:

x = some unknown number

2x = twice the number

(2x)^3 = cubing the previous result

If you want to simplify this expression, then you would get 8x^3 since 2^3 = 8.

4 0

2 years ago