All categories

3 years ago

8x-5y= 14 – 5х+бу = -3

Mathematics

1 answer:

podryga [215]3 years ago

4 0

Answer:

y= 34/23

x= 123/46

Step-by-step explanation:

Solve for x

8x-5y= 14

x= 14/8 +5/8y

Sub x into second equation

– 5х+бу = -3

-5(14/8 +5/8y) + 6y= -3

-29/4 +23/8y = -3

y= 34/23

Sub y into any equation and solve for x

8x=483/23

x= 123/46

You might be interested in

If the bookstore pays $60 to the publisher what will be the selling price?

Dmitriy789 [7]

Answer:

This means that, when the price of a book from a publisher is $60, the bookstore will sell it for $88 to the students.

4 0

3 years ago

What is 6/4 + 2/3 ? Im not very good at math so I need some help

makvit [3.9K]

Step-by-step explanation:

6/4 + 2/3 (taking LCM between denominators)

=> 6*3/4*3 + 2*4/3*4

=> 18/12 + 8/12

=> (18 + 8)/12

=> 26/12

=> 13/6

4 0

3 years ago

What is 85993 rounded to the nearest thousand

dsp73

86000 is the answer to the question

5 0

4 years ago

Read 2 more answers

If it is known that a+b/a-b=c, find the following: 2a-2b/5a+5b

Olin [163]

Answer:

$\frac{2a-2b}{5a+5b} =\frac{2}{5c}$

Step-by-step explanation:

$\frac{a+b}{a-b} =c$

$\Longleftrightarrow \frac{a-b}{a+b} =\frac{1}{c}$

$\Longleftrightarrow \frac{2}{5} \times \frac{a-b}{a+b} =\frac{2}{5} \times \frac{1}{c}$

$\Longleftrightarrow \frac{2\times(a-b)}{5\times(a+b)} = \frac{2\times1}{5\times c}$

$\Longrightarrow \frac{2a-2b}{5a+5b} =\frac{2}{5c}$

4 0

2 years ago

The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour. Ro

QveST [7]

Answer:

a) There is a 16.0623% probability that 5 messages are received in 1 hour.

b) There is a 11.5880% probability that 10 messages are received in 1.5 hours.

c) There is a 22.4042% probability that 2 messages are received in 0.5 hours.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have a mean of 6 messages per hour.

(a) What is the probability that 5 messages are received in 1 hour?

Find the value of P when $x = 5$ and $\mu = 6$

$P(X = 5) = \frac{e^{-6}*(6)^{5}}{(5)!} = 0.160623$

There is a 16.0623% probability that 5 messages are received in 1 hour.

(b) What is the probability that 10 messages are received in 1.5 hours?

The mean is 6 messages in one hour.

For 1.5 hours, the mean is 6*1.5 = 9 messages.

We have to find the value of P when $x = 10$ and $\mu = 9$ .

$P(X = 10) = \frac{e^{-9}*(9)^{10}}{(10)!} = 0.115880$

There is a 11.5880% probability that 10 messages are received in 1.5 hours.

(c) What is the probability that less than 2 messages are received in 1/2 hour?

The mean is 6 messages in one hour.

For 0.5 hours, the mean is 6*0.5 = 3 messages.

We have to find the value of P when $x = 2$ and $\mu = 3$ .

$P(X = 10) = \frac{e^{-3}*(3)^{2}}{(2)!} = 0.224042$

There is a 22.4042% probability that 2 messages are received in 0.5 hours.

6 0

3 years ago