If Sam's present age is x years, what will be his age in years after 20 years from now?

What are the center and radius f the circle with equation xthe power of 2 +y power of 2 -4x -6y +8=0

Dovator [93]

x² +y² -4x -6y +8=0
(x² -4x) +(y² -6y) = -8
We are going to complete square for x and y,
it should look like a²+2ab+b² = (a +b)², or a²-2ab+b² = (a -b)²,

(x²-2*2x+2²) -2² + (y²-2*3y +3²)-3²=- 8
(x-2)²+(y-3)²-4-9=-8
(x-2)²+(y-3)²=-8+13
(x-2)²+(y-3)²=5

Formula a circle (x-h)²+(y-k)²=R², where vertex has coordinates (h, k)

So , for our circle vertex (2,3) and radius = √5

6 0

4 years ago

Determine whether A = B, ACB, BCA, ACB, BCA, or if none of these applies.

spin [16.1K]

The relation between A and B subset is A⊆B

What is Subsets?

A set "X" is said to be the subset of "Y" is each elements of X is present in Y and is denoted by X⊆Y.

It should be noted that In roster notation, the use of the {" and "} character indicates that a set of numbers is a collection of elements, which can either be a finite or infinite collection.

Thus in the given question :

A={summer, autumn, spring, winter}

B={summer, winter}

"summer" and "winter" elements of B which is common and also presents in A therefore, B is subset of A and is denoted as A⊆B.

check and know more about sets here :

brainly.com/question/23454979

#SPJ1

4 0

1 year ago

Whats the is distance between (-9, 1) and (-9, 6)

Nadya [2.5K]

4 cubic squares is the answer

6 0

3 years ago

Read 2 more answers

Indicate which two quadrants “theta” could terminate if sin theta = 4/5

Nitella [24]

Answer:

1st and 2nd quadrants

Step-by-step explanation:

You have sine is positive since it is 4/5.

Sine is the y-coordinate.

On the coordinate plane y is positive in the 1st and 2nd quadrants.

8 0

3 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago