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3 years ago

3x - 5y = 14 and – 2x + 2y = 0

Mathematics

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

x=5.6

y=-5.6

Step-by-step explanation:

3x - 5y = 14 Equation 1

– 2x + 2y = 0 Equation 2

Simultaneous equation can be solved either through elimination method or substitution method. But we use elimination method for this question

Multiply equation 1 by -2 (the coefficient of x in equation 2) and multiply equation 2 with 3 (the coefficient of x in equation 1), so that x will have the same coefficient in the new equations, and easy to eliminate

-2(3x - 5y = 14)

-6x+10y=-28 Equation 3

3(– 2x + 2y = 0)

-6x+6y=0 Equation 4

Subtract equation 4 from 3 to eliminate x

-6x+10y=-28

-6x+6y=0

-6x-(-6x)=-6x+6x=0

10y-6y=5y

-28-0=-28

5y=-28

y=-28/5

y=-5.6

Substitute for y in equation 2

– 2x + 2y = 0

– 2x + 2(-5.6) = 0

-2x-11.2=0

-2x=11.2

x=-11.2/2

x=5.6

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At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa

Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation

n=580 represent the random sample taken

X represent the seafood sold in the country that is mislabeled or misidentified by the people

$\hat p=0.25$ estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

$\alpha=0.01$ represent the significance level (no given, but is assumed)

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204$

$0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296$

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

Part c

A government spokesperson claimed that the sample size was too small, relative to the billions of pieces of seafood sold each year, to generalize. Is this criticism valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

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3 years ago

What is an equation of the line that passes through the points (8,-8) and (-5,-8)

Vanyuwa [196]

Step-by-step explanation:

steps are in picture above.

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4 0

3 years ago

The average capacity of five containers is 13 liters. A container with a capacity of 7 liters is added to the set of 5. What is

Zepler [3.9K]

Answer:

12 liters

Step-by-step explanation:

Let the capacity sum of the 5 containers be X

13 = X / 5

X = 65 liters

Average capacity of the six = ( 65 + 7 )/ 6 liters

= 72/6 liters

= 12 liters

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3 years ago

Find the gradient of the line segment between the points (2,2) and (0,6).

Alexeev081 [22]

Answer:

slope-intercept formula -----> Y2-Y1/X2-X1

Step-by-step explanation:

X1 = 2 X2= 0

Y1 = 2 Y2=6

6-2/0-2= -2

G= -2

5 0

3 years ago