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kupik [55]
3 years ago
10

What is the solution to the system of linear equations?

Mathematics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

(0,2)

Step-by-step explanation:

The solution would just be where these two functions cross, which is at the point (0,2) in this case.

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I REALLY NEED HELP PLEASEEE!!! I'LL GIVE BRAINLIEST! PLEASE HELP!
marishachu [46]

Answer:

I think the answer would be B.

4 0
3 years ago
HELP PLEASE
kifflom [539]

Answer:

2nd choice

Step-by-step explanation:

to find the inital peanuts, we multiply 32 ounces by 3/10 or 0.3. then we add x which are peanuts as well. that makes 0.3 x 32 + x. this is our total peanuts. to find the percantage, we find the whole mixture which is now 32 + x, divide peanuts by the whole mixture and multiply by 100.

5 0
3 years ago
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The statistics of nequals22 and sequals14.3 result in this​ 95% confidence interval estimate of sigma​: 11.0less thansigmaless t
dimulka [17.4K]
Answer: no, the confidence interval for the standard deviation σ cannot be expressed as 15.7 \pm 4.7

There are three ways in which you can possibly express a confidence interval:

1) inequality
The two extremities of the interval will be each on one side of the "less then" symbol (the smallest on the left, the biggest on the right) and the symbol for the standard deviation (σ) will be in the middle:
11.0 < σ < 20.4
This is the first interval given in the question and it means <span>that the standard deviation can vary between 11.0 and 20.4

2) interval
</span>The two extremities will be inside a couple of round parenthesis, separated by a comma, always <span>the smallest on the left and the biggest on the right:
(11.0, 20.4)
This is the second interval given in the question.

3) point estimate </span><span>\pm margin of error</span>
This is the most common way to write a confidence interval because it shows straightforwardly some important information. 
However, this way can be used only for the confidence interval of the mean or of the popuation, not for he confidence interval of the variance or of the standard deviation.

This is due to the fact that in order to calculate the confidence interval of the standard variation (and similarly of the variance), you need to apply the formula:
\sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{\alpha / 2} } } \leq \sigma \leq \sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{1 - \alpha / 2} } }

which involves a χ² distribution, which is not a symmetric function. For this reason, the confidence interval we obtain is not symmetric around the point estimate and the third option cannot be used to express it.
4 0
3 years ago
Can you help with this?
gregori [183]
The slope of the given line  = 1  so the slope of the line perpendicular to it will be -(1/1)  = -1.
It also passes through point (1, -1)  so we use the point slope equation

y - y1 = m(x - x1)

y - -1) = -1(x - 1)
y + 1 = -x + 1
y = - x    <--------- that's the answer


4 0
3 years ago
ASAP! I really need help with this question! No nonsense answers, and please attach the solution.
uysha [10]

Answer:

\boxed{\sf Option \ 4}

Step-by-step explanation:

\sqrt{2x-3} +x=3

Subtract x from both sides.

\sqrt{2x-3} +x-x=-x+3

\sqrt{2x-3}=-x+3

Square both sides.

( \sqrt{2x-3})^2 =(-x+3)^2

2x-3=x^2-6x+9

Subtract x²-6x+9 from both sides.

2x-3-(x^2-6x+9 )=x^2-6x+9-(x^2-6x+9)

-x^2 +8x-12=0

Factor left side of the equation.

(-x+2)(x-6)=0

Set factors equal to 0.

-x+2=0\\-x=-2\\x=2

x-6=0\\x=6

Check if the solutions are extraneous or not.

Plug x as 2.

\sqrt{2(2)-3} +2=3\\ \sqrt{4-3} +2=3\\\sqrt{1} +2=3\\3=3

x = 2 works in the equation.

Plug x as 6.

\sqrt{2(6)-3} +6=3\\ \sqrt{12-3} +6=3\\\sqrt{9} +6=3\\3+6=3\\9=3

x = 6 does not work in the equation.

5 0
3 years ago
Read 2 more answers
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