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Alex
3 years ago
14

Create a system of equations and use algebra to write a quadratic function that passes through the

Mathematics
1 answer:
skad [1K]3 years ago
6 0

y = x² + 3x + 10

the equation of a parabola in standard form is → y = ax² + bx + c ( a ≠ 0 )

Substitute each of the pairs of coordinate points into the equation and solve for a, b and c

(- 2, 8) → 8 = 4a + 2b + c → (1)

(1, 14) → 14 = a + b + c → (2)

note that (0 , 10) is the y-intercept ⇒ c = 10

substitute c = 10 into (1) and (2)

8 = 4a - 2b + 10 → 4a - 2b = - 2 → (3)

14 = a + b + 10 → a + b = 4 → (4)

multiply (4 ) by 2

2a + 2b = 8 → (5)

add (3) and (5) term by term ⇒ 6a = 6 ⇒ a = 1

substitute a = 1 into (4)

1 + b = 4 ⇒ b = 3

thus a = 1, b= 3 and c = 10

equation of parabola is y = x² + 3x + 10



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a_n=2a_{n-1}+1
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What is an equation of the line that passes through the points (-4,-2)<br> and (8, 1)?
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y=1/4x-1

Step-by-step explanation:

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3 years ago
When consumers apply for credit, their credit is rated using FICO (Fair, Isaac, and Company) scores. Credit ratings are given be
Flura [38]

Answer:

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

Step-by-step explanation:

Previous concepts

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 data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.  

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n}  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}  

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is \bar X= 660.313

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. The degrees of freedom are given by:

df=n-1=16-1=15

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for t_{\alpha/2}=-2.95

"=T.INV(1-0.005,15)" for t_{1-\alpha/2}=2.95

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}  

And if we find the limits we got:

660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588  

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]  

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

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the 3rd one. hope this helps

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Vladimir [108]

Answer:

120th caller

Step-by-step explanation:

120 is the common number by 40 and 30. trust me on this I'm really good at this subject in math.

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