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3 years ago

Find an equation for the circle that has center (−3, 4) and passes through the point (4, −2).

1 answer:

dsp733 years ago

8 0

First use the coordinate of the center to start building your equation. The equation of a circle recall is
${(x - h)}^{2} + {(y - k)}^{2} = {r}^{2}$
With the center point being (h,k).

Once you put in the centers coordinates, your equation should be,
${(x + 3)}^{2} + {(y - 4)}^{2} = {r}^{2}$
However we still need to find the radius. We can do this using the second point given, by putting it into the equation and solving for the radius.
${( 4 + 3)}^{2} + {( - 2 - 4)}^{2} = {r}^{2}$
Solving this gives us the final part of the equation.
$85 = {r}^{2}$
And we can finally put our full equation together,
${(x + 3)}^{2} + {(y - 4)}^{2} = 85$

You might be interested in

An author argued that more basketball players have birthdates in the months immediately following July 31, because that was the

Lady_Fox [76]

Answer:

There is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same frequency.

Step-by-step explanation:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same frequency.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

$\chi^{2}=\sum{\frac{(O-E)^{2}}{E}}$

The values are computed in the table.

The test statistic value is $\chi^{2}=128.12$ .

The degrees of freedom of the test is:

n - 1 = 12 - 1 = 11

Compute the p-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same frequency.

6 0

3 years ago

What is the vaule of 0.77

padilas [110]

seventy-seven hundredths.

3 0

3 years ago

Suppose you can work at most a total of 25 hours per week. Baby-sitting, x, pays $6 per hour and working at the grocery store, y

Andrei [34K]

Answer:

y=7 number of hours at grocery store

x=18 number of hours at baby- sitting

Step-by-step explanation:

According to the information provided.

x is number of hours at baby- sitting

y is number of hours at grocery store

total number of hours worked

1) x+y =25

total earn in a week

2) x*$6 + y* $9 = $171

from equation 1

x+y=25

x= 25-y

we place the above derived equation in equation 2

x*$6 + y* $9 = $171

(25-y)*$6 + y* $9 = $171

(25*6) -6y +9y =171

150+3y=171

3y=171-150

3y=21

y=7 number of hours at grocery store

x= 25-y

x= 25-7

x=18 number of hours at baby- sitting

3 0

3 years ago

Which equation describes the same line as y- 5 = -2(x + 4)?

lord [1]

Y-5=-2(x+4)
Y-5=-2x-8
Add 5 to both sides
Y=-2x-3
D

8 0

3 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago