Please help me idk this

PLEASE HELP ILL GIVE BRAINLIEST

DanielleElmas [232]

Option A:

$(x-5)^{2}+(y-3)^{2}=16$

Solution:

Given data:

Center of the circle is (5, 3).

Radius of the circle = 4

To find the equation of the circle:

The general form of the equation of a circle in centre-radius format is

$(x-h)^{2}+(y-k)^{2}=r^{2}$

where (h, k) is the centre of the circle and r is the radius of the circle.

Substitute the given values in the equation of a circle formula:

$(x-5)^{2}+(y-3)^{2}=4^{2}$

$(x-5)^{2}+(y-3)^{2}=16$

The equation of the given circle is $(x-5)^{2}+(y-3)^{2}=16$ .

Hence Option A is the correct answer.

5 0

3 years ago

Jane had of a meter of ribbon. She used of a meter of ribbon to decorate a card. How much ribbon did she have left after she dec

emmainna [20.7K]

She had none left. if she only had a meter and then used a meter there would be none left.

5 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

Use an equivalent fraction to find the sum of 4/10 + 28/100

kipiarov [429]

Hi there!

4/10 = (4*10)/(10*10) = 40/100

40/100 + 28/100 = 68/100

68/100 = 34/50 = 17/25

Hope this helps!

4 0

3 years ago

Read 2 more answers

Solve the equation. 5x + 8 - 3x = -10 x = -1 x=1 x=9

aksik [14]

Answer:

solution,

 $5x + 8 - 3x = - 10 \\ or \: 5x - 3x + 8 = -10 \\ or \: 2x + 8 = -10 \\ or \: 2x = -10 - 8 \\ or \: 2x = -18\\ or \: x = \frac{-18}{2 } \\ x = -9$ 

hope this helps..

Good luck on your assignment

7 0

4 years ago

Read 2 more answers