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PilotLPTM [1.2K]

3 years ago

15

The point (1,4) lies on a circle that is centered at (1, 1). Which statements are correct?

1 answer:

xenn [34]3 years ago

6 0

Answer:

The circle's radius is 3 units.

Step-by-step explanation:

Let's get to the answer using the formula for EQUATION OF A LINE at the center of a circle.

The equation is expressed as:

(x - h) ^2 +. (y - k) ^ 2 = r ^2

Where:

(h, k) = The center of the circle

(x, y) = Some points around the circle

(r) = Radius of the circle

Let's apply the formula to solve the question:

(x - h) ^2 +. (y - k) ^ 2 = r ^2

(1 - 1)^2 + (4 - 1)^2 = r^2

0^2 + 3^2 = r^2

0 + 9 = r^2

r^2 = 9

r = √9

r = 3

Therefore,

The circle's radius is 3 units.

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Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Find the percent of change from 24°F to 39°F. Then state whether the percent of change is an increase or a decrease.

lara31 [8.8K]

Answer:

24F is a increase I belive and 39F is a decrease

4 0

3 years ago

Compute the requested value. Choose the correct answer.

Fiesta28 [93]

Original price of the item = $14.95

Price after discount = $13.79

Discount offered = original price - price after discount = 14.95- 13.79 = $1.16

Now let us find the percentage of discount offered.

Percentage discount is given by the formula:

$Percentage discount = \frac{MP-SP}{MP}*100$

Where MP= Marked price= original price

SP= selling price= price after discount

$Percentage discount = \frac{1.16}{14.95}*100$

Percentage discount = 7.759 %

7 0

3 years ago

Read 2 more answers

Let A(t) be the area of a circle with radius r(t), at time t in min. Suppose the radius is changing at the rate of drdt=6 ft/min

mylen [45]

Answer:

The rate of change is $108\pi$ ft^(2)/min

Step-by-step explanation:

The area of a circle is given by the following equation:

$A(t) = \pi r^{2}$

To solve this question, we have to realize the implicit differentiation in function of t. We have two variables, A and r. So

$\frac{dA(t)}{dt} = 2\pi r \frac{dr}{dt}$

We have that:

$\frac{dr}{dt} = 6, r = 9$ .

We want to find $\frac{dA}{dt}$

So

$\frac{dA(t)}{dt} = 2\pi*9*6$

$\frac{dA}{dt} = 108\pi$

Since the area is in square feet, the rate of change is in ft^(2)/min.

So the rate of change is $108\pi$ ft^(2)/min

5 0

3 years ago

Can I get help with this?

blagie [28]

Answer:

y-4=7(x-1)

Step-by-step explanation:

Hi there!

We are given the slope of the line (7) and the point (1,4) and we need to find the equation in point-slope form.

Point-slope form is given as y- $y_{1}$ =m(x- $x_{1}$ ) where m is the slope and ( $x_{1}$ , $y_{1}$ ) is a point

We have all of the needed information for the equation, but let's first label the values of everything in order to avoid confusion

m=7

$x_{1}$ =1

$y_{1}$ =4

now substitute into the formula:

<u>y-4=7(x-1)</u>

Hope this helps! :)

6 0

2 years ago