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3 years ago

What is X and Y ? Please as fast as you can answer this

Mathematics

1 answer:

olganol [36]3 years ago

7 0

Answer:

28/5=x and y= -6/5

Step-by-step explanation:

solve by substitution

3x+4y=12

Rewrite equations:

2x+y=10;3x+4y=12

Step1: Solve 2x+y=10 for y:

2x+y=10

2x+y+−2x=10+−2x(Add -2x to both sides)

y=−2x+10

Step2: Substitute−2x+10 for y in 3x+4y=12:

3x+4y=12

3x+4(−2x+10)=12

−5x+40=12(Simplify both sides of the equation)

−5x+40+−40=12+−40(Add -40 to both sides)

−5x=−28

-5x/-5=-28/5

(Divide both sides by -5)

x= 28/5

Step3: Substitute

28/5 for x in "y=−2x+10":

y=−2x+10

y=−2(28/5)+10

y=-6/5(Simplify both sides of the equation)

Hope this helps!

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Answer:

Mean: 11.875 or 11.9

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Step-by-step explanation:

Mean = all numbers added up and then divided by the amount of numbers

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Step-by-step explanation:

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3 years ago

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) 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 t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

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 t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

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Answer:

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Step-by-step explanation:

ur welcome

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Answer:

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Step-by-step explanation:aw dawd awd a

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3 years ago

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