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3 years ago

Solve the system of equations. -12x-5y=40 and 12x-11y=88 x= y=

Mathematics

2 answers:

SpyIntel [72]3 years ago

3 0

Answer:

(0,-8)

x=0

y=-8

Step-by-step explanation:

There are a couple of ways you can solve this I used what I think is called the elimination method

We first start by writing the equations in linear form (y=mx+b)

For the first one we have -12x-5y=40

we add 5y to both sides and subtract 40 to get

-12x-40=5y

divide by 5 to get

-12/5x-8=y

Then the second one we have

12x-11y=88

same deal here, add 11y and subtract 88 to get

12x-88=11y

divide by 11 and get

12/11x-8=y

We then set the two equations equal to each other and get

-12/5x-8=12/11x-8

We can add 8 to both sides and get

-12/5x=12/11x

Just by looking at it you can see that the only solution for x is 0

Plug in 0 to the first equation and get -12*0-5y=40

This means that -5y=40

Solve for y and get -8

Aleksandr-060686 [28]3 years ago

3 0

Answer:

x=0 , y= -8

Step-by-step explanation:

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In a sequnece u5=-3.7 and u15 =-52.3, what is the 19th term

PilotLPTM [1.2K]

u19=-71.74

Step-by-step explanation:

u5=a+4d=-3.7....(1)

u15=a+14d=-52.3....(2)

-10d=48.6

d=-48.6/-10

d=-4.86

Substitute-4.86 into....(1)

u5=a+4(-4.86)=-3.7

a+(-19.44)=-4.7

a=-3.7-(-19.44)

a=15.74

19term=a+18d=?

=15.74+18(-4.86)

=15.74+(-87.48)

19th term =-71.74

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2 years ago

If the graph of the equation y=(x+2)^2 is reflected with respect to the y-axis, what is the equation of the resulting graph?

alexandr1967 [171]

Answer:

Is important to remember that the reflection of the point (a,b) across the y-axis is the point (-a,b).

Our original equation is $y= (x+2)^2$

And after the reflection respect x we got:

$y= (-x+2)^2$

Step-by-step explanation:

Is important to remember that the reflection of the point (a,b) across the y-axis is the point (-a,b).

Our original equation is $y= (x+2)^2$

And after the reflection respect x we got:

$y= (-x+2)^2$

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sergiy2304 [10]

Answer:

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Find the distance from point A to the given line<br> A(2,-1), y=-x+4

BARSIC [14]

Answer:

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3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

3 years ago