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1 year ago

4x + 5y = 3 2x + 3y = 1Solve by elimination

Mathematics

1 answer:

Lesechka [4]1 year ago

7 0

Answer:

The solution to the system of equations is:

x = 2, and y = -1

Explanation:

Given the pair of equations:

4x + 5y = 3 ..........................................................................(1)

2x + 3y = 1............................................................................(2)

To solve this by elimination:

Multiply equation (2) by 2, to eliminate x

Equation (2) becomes

4x + 6y = 2 .........................................................................(3)

Subtract equation (1) from (3)

4x - 4x + 6y - 5y = 2 - 3

y = -1 ....................................................................................(4)

Multiply equation (1) by 3 and equation (2) by 5 to eliminate y

Equation (1) becomes

12x + 15y = 9 .......................................................................(5)

Equation (2) becomes

10x + 15y = 5 ........................................................................(6)

Subtract equation (6) from (5)

12x - 10x + 15y - 15y = 9 - 5

2x = 4

Divide both sides by 2

x = 4/2 = 2 ............................................................................(7)

From equations (7) and (4)

x = 2, and y = -1

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3 years ago

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In the first diagram the value of k is 10 and in the second diagram the value of k is 15/2.

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The triangle can be defined as a three-sided polygon in geometry, and it consists of three vertices and three edges. The sum of all the angles inside the triangle is 180°.

In the first diagram:

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2 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

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