1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ahrayia [7]
3 years ago
9

Solve the system of equations. 2x+8y=6 -5x-20y+-15

Mathematics
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer:

2x+8y=6

Step 1: Add -8y to both sides.

2x+8y+−8y=6+−8y

2x=−8y+6

Step 2: Divide both sides by 2

\frac{2x}{2} = \frac{-8y+6}{2}

x=−4y+3

_________________________________________________________

−5x−20y−15

There are no like terms.

Answer:

=−5x−20y−15

You might be interested in
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
2 years ago
Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
zloy xaker [14]

Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

4 0
3 years ago
Please help me to evaluate 6+7=8−15
zzz [600]
13=-7 which is not true
3 0
3 years ago
Read 2 more answers
A number is greater than 50 and less than 85 the number has 13 and 3 as factors
alina1380 [7]

Answer:

The Answer is: 78.

Step-by-step explanation:

This one makes you think!

Let n = the number.

The number is greater than 50 and less than 85:

50 < n < 85

Some of the factors are 13 and 3.

So, multiply 13 * 3 = 39.

Multiply again by 2:

39 * 2 = 78.

78 can be factored as:

13 * 3 * 2 = 78, and it is greater than 50 and less than 85.

The number is 78.

Hope this helps! Have an Awesome Day!! :-)

8 0
3 years ago
the first term of a geometric progression is 12 and the second term is -6. Find the (i) the tenth term of the progression (ii) t
lesantik [10]

Step-by-step explanation:

a1=ar⁰=12 A2=ar¹=-6

a=12 ar=-6

12r=-6

r=-6/12

r=-1/2

a10=ar⁹

=12*(-1/2)⁹

=12*(-512)

=-3/128

S∞ = a1 / (1-r )

=12/(1-1/2)

=12*1/2

=6

6 0
3 years ago
Other questions:
  • P&gt;-1 I need help fast I have to pass this
    14·1 answer
  • I need help setting the equation up??
    13·1 answer
  • How would i graph this?<br><br>someone help me please
    13·1 answer
  • Solve 5x – 2y = 11 for y.​
    15·1 answer
  • Which of the following is NOT a principle of making inferences from dependent​ samples? Choose the correct answer below. A. The
    15·1 answer
  • PLEASSSSSEEE HELP A GIRL OUT PLZZ​
    15·1 answer
  • Please please help please please ASAP please I'm begging you please I have 10 minutes please y'all help please any ACE that can
    5·2 answers
  • ANSWERS QUICKLY PLEASE
    6·1 answer
  • HELPPP ME PLSSSS<br> I NEED THIS FOR MY CLASS
    14·1 answer
  • -3(7+6t)<br> Can you tell me the answer
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!