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Komok [63]
2 years ago
11

What is the length of the conjugate axis? (y-2)^2/16 -(x+1)^2/144=1

Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
7 0

Answer: 24

A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point in the same plane to its distance from a fixed line is always constant, which is always greater than unity.

the fixed point is called the focus and the fixed line is directrix and the ratio is the eccentricity.

The general equation for the vertical hyperbola is

            [ (y-k)^2 / a^2 ] – [ (x-h)^2 / b^2 ] = 1

The conjugate axis of the vertical hyperbola is y = k

Length of the conjugate axis  = 2b

According to the question k = 2, h = -1, a = 4, b = 12

Length of the conjugate axis = 2b = 2 * 12  = 24

Learn more about hyperbola here :

brainly.com/question/3351710

#SPJ9

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Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x
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Answer:

a) Figure attached

b) y=1.31 x +98.57

c) The correlation coefficient would be r =0.47719

d) y=1.31 x +98.57=1.31*21 + 98.57 =126.08

Step-by-step explanation:

(a) Draw a scatter diagram for the data.

See the figure attached

(b) Find x, y, b, and the equation of the least-squares line. (Round your answers to three decimal places.) x =__ y =__ b =__ y =__ + __x

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6576-\frac{300^2}{14}=147.429

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=38186-\frac{300*1773}{14}=193.143

And the slope would be:

m=\frac{193.143}{147.429}=1.31

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{300}{14}=21.429

\bar y= \frac{\sum y_i}{n}=\frac{1773}{14}=126.643

And we can find the intercept using this:

b=\bar y -m \bar x=126.643-(1.31*21.429)=98.571

So the line would be given by:

y=1.31 x +98.57

(c) Find the sample correlation coefficient r and the coefficient of determination r?2. (Round your answers to three decimal places.)

n=14 \sum x = 300, \sum y = 1773, \sum xy=38186, \sum x^2 =6576, \sum y^2 =225649  

And in order to calculate the correlation coefficient we can use this formula:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}

r=\frac{14(38186)-(300)(1773)}{\sqrt{[14(6576) -(300)^2][14(225649) -(1773)^2]}}=0.9534

So then the correlation coefficient would be r =0.47719

What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.)

The % of variation is given by the determination coefficient given by r^2 and on this case 0.47719^2 =0.2277, so then the % of variation explaines is 22.8%.

(d) If a female baby weighs 21 pounds at 1 year, what do you predict she will weigh at 30 years of age? (Round your answer to two decimal places.) ___ lb

So we can replace in the linear model like this:

y=1.31 x +98.57=1.31*21 + 98.57 =126.08

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Recall the binomial theorem.

(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k

1. The binomial expansion of \left(1+\frac x3\right)^7 is

\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}

Observe that

k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x

k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2

When we multiply these by 8-9x,

• 8 and \frac73 x^2 combine to make \frac{56}3 x^2

• -9x and \frac73 x combine to make -\frac{63}3 x^2 = -21x^2

and the sum of these terms is

\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}

2. The binomial expansion is

\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k

We get the a^6b^2 term when k=2 :

k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2

6 0
1 year ago
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