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3 years ago

Find the Center and the radius of the circle x2 + y2 - 6x + 8y- 7 = 0.

Mathematics

1 answer:

777dan777 [17]3 years ago

8 0

Answer:

C(3,-4), r=4*sqrt(2)

Step-by-step explanation:

C(p, q)

x^2+y^2+dx+ey+f=0

p=-d/2, q=-e/2, r^2=p^2+q^2-f

x^2+y^2 - 6x+8y-7=0

p=-(-6)/2=6/2=3

q=-8/2=-4

r^2=3^2 +(-4)^2+7

r^2=9+16+7

r^2=32

r=sqrt(32)

r=sqrt(16*2)

r=4*sqrt(2)

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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy

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Answer:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

$df = n-1=75-1 =74$

The p value is:

$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

$\bar X = 3.13 , s =0.03 , n =75$

The statistic to check the hypothesis is given by:

$t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

Replacing the info given we got:

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

The degrees of freedom are given by:

$df = n-1=75-1 =74$

The p value since is a right tailed ted is given by:

$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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You bike 1 mile the first day of your training, 1.3 miles the second day, 1.9 miles the third day, and 3.1 miles the fourth d

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Answer:

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Step-by-step explanation:

In this problem we have:

$d_1=1 mi$ is the distance travelled during the 1st day

$d_2=1.3 mi$ is the distance travelled during the 2nd day

$d_3=1.9 mi$ is the distance travelled during the 3rd day

$d_4=3.1 mi$ is the distance travelled during the 4th day

We notice that the difference between the distance travelled on the (n+1)-th day and the distance travelled on the n-th day doubles every day. In fact:

$d_2-d_1=0.3\\d_3-d_2=2\cdot 0.3 = 0.6\\d_4-d_3=2\cdot 0.6 = 1.2$