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3 years ago

Write the equation of the circle that has its center at (4, 5) and a radius of 3.

Mathematics

1 answer:

alexdok [17]3 years ago

5 0

Answer:

$(x-4)^{2} +(y-5)^{2} =9$

Step-by-step explanation:

The equation for a circle is $(x-h)^{2} +(y-k)^{2} =r^{2}$

(h, k) is the center

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Find the sum of 2x2 3x - 4, 8 - 3x, and -5x2 2. -3x2 - 2 -7x2 - 2 -3x2 6 -7x2 - 6x - 2

Law Incorporation [45]

-3x to the second power - 2

I believe this is your answer.

4 0

3 years ago

Is The polar form of a complex number is unique?

Pie

Answer:

Step-by-step explanation:

The angle in the polar form of a complex number can have any multiple of 2π radians added to it, and the number will be the same number. That is, there are an infinite number of representations of a complex number in polar form.

6 0

3 years ago

Circle A is inscribed in a quadrilateral. What is the perimeter of the quadrilateral?

faust18 [17]

AB + CD = BC + DA
<span>Where A, B, C, and D are points on the quad, and AB is the length of the line A to B, etc. </span>

<span>Since you know three sides, you can easily solve for the fourth. </span>

6 0

3 years ago

3. A new process has been developed for applying pho toresist to 125-mm silicon wafers used in manufac turing integrated circu

dlinn [17]

Answer:

The null hypothesis is $H_o : \mu = 13.4000$

The alternative hypothesis is $H_a : \mu \ne 13.4000$

The null hypothesis is rejected

The 99% confidence level is $13.3930 < \mu < 13.3994$

Step-by-step explanation:

From the question we are told that

The sample size is n = 10

The population mean is $\mu = 13.4 000 \ angstroms$

The level of significance is $\alpha = 0.05$

The sample data is

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

$\= x = \frac{13.3987+ 13.3957\cdots +13.4002 }{10}$

=> $\= x = 13.3962$

Generally the sample standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }$

=> $\sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 + (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2 }{10} }$

=> $\sigma =0.0039$

The null hypothesis is $H_o : \mu = 13.4000$

The alternative hypothesis is $H_a : \mu \ne 13.4000$

Generally the test statistics is mathematically represented as

$z = \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }$

=> $z = \frac{13.3962 - 13.4000}{\frac{0.0039}{\sqrt{10} } }$

=> $z = 3.08$

Generally the p-value is mathematically represented as

$p-value = 2P( z > 3.08)$

From the z-table $P(z > 3.08)= 0.001035$

$p-value = 2* 0.001035$

$p-value = 0.00207$

So from the obtained value we see that

$p-value < \alpha$

Hence the null hypothesis is rejected

Consider the b question

Given that the confidence level is 99% then the level of significance is

$\alpha = (100 -99)\%$

=> $\alpha = 0.01$

Generally from the normal distribution table critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 2.58 * \frac{0.0039}{\sqrt{10} }$

=> $E = 0.00318$

Generally the 99% confidence interval is mathematically represented as

$13.3962 - 0.00318 < \mu < 13.3962 + 0.00318$

=> $13.3930 < \mu < 13.3994$

7 0

3 years ago

Suppose a random sample of 50 college students are asked to measure the length of their right foot in centimeters. A 95% confide

hichkok12 [17]

Answer:

A 99% confidence interval will be wider than a 95% confidence interval

Step-by-step explanation:

From the question we are told that

The 95% confidence interval for for the mean foot length for students at the college is found to be 21.709 to 25.091 cm

Generally the width of a confidence interval is dependent on the margin of error.

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

From the above equation we see that

$E \ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Here $Z_{\frac{\alpha }{2} }$ is the critical value of the half of the level of significance and this value increase as the confidence level increase

Now if a 99% confidence level is used , it then means that the value of

$Z_{\frac{\alpha }{2} }$ will increase, this in turn will increase the margin of error and in turn this will increase the width of the confidence interval

Hence a 99% confidence interval will be wider than a 95% confidence interval

5 0

3 years ago