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3 years ago

Simultaneous equations x^2+3xy=10 x=2y

Mathematics

1 answer:

satela [25.4K]3 years ago

8 0

9514 1404 393

Answer:

(x, y) = (-2, -1) or (2, 1)

Step-by-step explanation:

Substitute for x in the first equation:

(2y)^2 +3(2y)y = 10

10y^2 = 10

y^2 = 1

y = ±1

x = 2y = ±2

Solutions are (x, y) = (-2, -1) or (2, 1).

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27 boiz

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3 years ago

Read 2 more answers

When network cards are communicating, bits can occasionally be corrupted in transmission. Engineers have de- termined that the n

ycow [4]

Answer:

a) 11.40% probability of 5 bits being in error during the transmission of 1 kb

b) 11.60% probability of 8 bits being in error during the transmission of 2 kb

c) 0.01% probability of no error bits in 3kb

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 giveninterval.

Poisson distribution with mean of 3.2 bits/kb (per kilobyte).

This means that $\mu = 3.2kb$ , in which kb is the number of kilobytes.

(a) What is the probability of 5 bits being in error during the transmission of 1 kb?

This is P(X = 5) when $\mu = 3.2$

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

$P(X = 5) = \frac{e^{-3.2}*3.2^{5}}{(5)!} = 0.1140$

11.40% probability of 5 bits being in error during the transmission of 1 kb

(b) What is the probability of 8 bits being in error during the transmission of 2 kb?

This is P(X = 8) when $\mu = 2*3.2 = 6.4$

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

$P(X = 8) = \frac{e^{-6.4}*6.4^{8}}{(8)!} = 0.1160$

11.60% probability of 8 bits being in error during the transmission of 2 kb

(c) What is the probability of no error bits in 3kb?

This is P(X = 0) when $\mu = 3*3.2 = 9.6$

Then

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

$P(X = 0) = \frac{e^{-9.6}*9.6^{0}}{(0)!} = 0.0001$

0.01% probability of no error bits in 3kb

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3 years ago

Prior to a special advertising campaign, 23% of all adults recognized a particular companyâs logo. At the close of the campaign

pochemuha

Answer:

The answer is "2.4049"

Step-by-step explanation:

Calculating the test of Hypothesis: $H_{0}: 23\% \ \text{off all adults which reconize the compony's logo}\\\\H_{1}: \text{more than 23\% of adult recornise the compony's logo}\\\\$

that is

$H_{0}: p=0.23\ against \ H_{1}:p>0.01\\\\Z=\frac{P-p}{\sqrt{\frac{p(1-p)}{n}}}\sim N(0,1)\\\\$

Given:

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Z=2.576 tabled value. Because Z is 2.4049, that's less than Z stated, there is no indication that a null hypothesis is rejectable, which means that 23% of all adults record the logo of the Company.

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3 years ago

During the exponential phase, E. coli bacteria in a culture increase in number at a rate proportional to the current population.

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Answer:

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Step-by-step explanation:

The appropriate exponential equation for the population is ...

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Then we can compute for t=7.2:

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7.2 minutes from now, the population will be about 197.2 million.