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3 years ago

2x-37=13 x+2y+-4 Linear Combination Help please

1 answer:

5 0

Solve the following system:
{2 x - 37 = 13 | (equation 1)
{x + 2 y = -4 | (equation 2)

Express the system in standard form:
{2 x+0 y = 50 | (equation 1)
{x + 2 y = -4 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:
{2 x+0 y = 50 | (equation 1)
{0 x+2 y = -29 | (equation 2)

Divide equation 1 by 2:
{x+0 y = 25 | (equation 1)
{0 x+2 y = -29 | (equation 2)

Divide equation 2 by 2:
{x+0 y = 25 | (equation 1)
{0 x+y = (-29)/2 | (equation 2)

Collect results:

Answer: {x = 25, y = -29/2

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What is .55, 0.59, .091, .0019, .0091, 5.0, 90.5 in order from least to greatest

postnew [5]

Answer:

0.0019, 0.0091, 0.091, 0.55, 0.59, 5.0, 90.5

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3 years ago

A 50 lb bag of grass seed cost $222.50 find the unit cost per lb of this grass seed

myrzilka [38]

To find the cost per one pound, we divide the total cost $222.50 by the total number of pounds (50 lb).

= $222.50 ÷ 50
= $4.45 per pound

ANSWER: Each pound of grass seed costs $4.45.

Hope this helps! :)

3 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

Find the Slope and the y- intercept for 5x+5y=-40

zheka24 [161]

Answer:

y-intercept= (0,-8)

slope= -1

Step-by-step explanation:

3 0

3 years ago

A pigeon can fly at a speed of 84 kilometers per hour. How long does it take the pigeon to fly 7 kilometers?

Serhud [2]

The pigeon can fly 7 kilometers in around 8-9 minutes

3 0

3 years ago