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Kay [80]
3 years ago
8

1. Is the function below linear or non linear? y = -8X2 + 9??​

Mathematics
1 answer:
romanna [79]3 years ago
3 0
Linear because it’s in y=mx+b form
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Please help me find m<1 before 11:59 PM!
Bond [772]

Answer:

129 degrees

Step-by-step explanation:

add the interior angles: 62 + 67 = 129

subtract your answer from 180 to get the exterior angle of the interior angle that is opposite to the given angles

4 0
3 years ago
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Find the slope for (3,4) and (3,-4)
saveliy_v [14]

Answer:

undefined

The problem:

Find the slope for the line going through (3,4) and (3,-4).

Step-by-step explanation:

Line up points vertically and subtract.

Then put 2nd difference over 1st.

( 3 ,  4)

-(3  , -4)

------------

0      8

So the slope would have been 8/0 but this is undefined.

So the slope is undefined.

Also notice the x's are the same and the y's are difference so this is a vertical line. There is only rise in a vertical line and no run.  Recall, slope is rise/run. You cannot divide by 0 so this is why we say the slope is undefined when the x's are always the same no matter the y.

7 0
3 years ago
At a restaurant, your bill has a food total of $45.50. Find the amount of an 18% tip on the food total.
oksano4ka [1.4K]
$53.69

45.50+8.19 (45.50 times 18%)=$53.69
8 0
3 years ago
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Hi! I was wondering if you could help with this question please :)​
Makovka662 [10]

Answer:

R=\frac{QJ}{I^2t}

Step-by-step explanation:

So we have the equation:

Q=\frac{I^2Rt}{J}

And we want to solve for R.

First, let's multiply both sides by J to remove the fraction on the right. So:

(J)Q=(J)\frac{I^2Rt}{J}

Simplify the right:

JQ=I^2Rt

We can rewrite our equation as:

JQ=R(I^2t)

So, to isolate the R variable, divide both sides by I²t:

\frac{JQ}{I^2t}=\frac{R(I^2t)}{I^2t}

The right side cancels, so:

R=\frac{QJ}{I^2t}

And we are done!

7 0
3 years ago
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
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