All categories

2 years ago

Answer!asap if you are a bot don't even respond

Mathematics

1 answer:

Tema [17]2 years ago

6 0

Answer:

Step-by-step explanation:

Upon factoring all terms we are left with the product of

[(x-4)(x+6)(x-6)]/[5(x-6)(3x+5)(x-4)]

The (x-4)s and (x-6)s cancel out and we are left with

(x+6)/(5(3x+5)) which is also equal to

(x+6)/(15x+25)

You might be interested in

Which students line segments have a midpoint of (-1,4)? Select all that apply A. Student 1 B. Student 2 C. Student 3 D. Student

SashulF [63]

Answer:

C. Student 3

E. Student 5

Step-by-step explanation:

we know that

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

Verify the midpoint of each student

student 1

we have the endpoints

(-9,0) and (11,-8)

substitute in the formula

$M(\frac{-9+11}{2},\frac{0-8}{2})$

$M(1,-4)$

The midpoint is not (-1,4)

student 2

we have the endpoints

(-6,-1) and (4,-7)

substitute in the formula

$M(\frac{-6+4}{2},\frac{-1-7}{2})$

$M(-1,-4)$

The midpoint is not (-1,4)

student 3

we have the endpoints

(-5,2) and (3,6)

substitute in the formula

$M(\frac{-5+3}{2},\frac{2+6}{2})$

$M(-1,4)$

The midpoint is equal to (-1,4)

student 4

we have the endpoints

(-3,10) and (5,-2)

substitute in the formula

$M(\frac{-3+5}{2},\frac{10-2}{2})$

$M(1,4)$

The midpoint is not (-1,4)

student 5

we have the endpoints

(0,-3) and (-2,11)

substitute in the formula

$M(\frac{0-2}{2},\frac{-3+11}{2})$

$M(-1,4)$

The midpoint is equal to (-1,4)

therefore

Student 3 and student 5

3 0

3 years ago

We would like to create a confidence interval.

Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level $1-\alpha$ , we have the confidence interval with a margin of error of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of $\sqrt{\pi(1-\pi)}$

a.A 99% confidence level and a sample size of 50 subjects.

$n = 50$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}$

b.A 90% confidence level and a sample size of 50 subjects.

$n = 50$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}$

c.A 90% confidence level and a sample size of 300 subjects.

$n = 300$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}$

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

$n = 300$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}$

6 0

3 years ago

Need help finding the area of the shaded region. Round to the nearest tenth. Need help ASAP

Sergio [31]

Answer: 294.4 m²

Step-by-step explanation:

Separate the shaded region into two parts:

The section containing the central angle of 230° (360° - 130°)
The triangle with sides 11.1, 11.1 & 20.12 (use Law of Cosines)

$1.\ Area(A)=\pi\ r^2\ \bigg(\dfrac{\theta}{360}\bigg)\\\\\\.\qquad \qquad =\pi(11.1)^2\bigg(\dfrac{230}{360}\bigg)\\\\\\.\qquad \qquad =247.3$

$2.\ \text{Use Law of cosines to find the length of the third side.}\\\text{ Then use Heron's formula to find the Area of the triangle.}\\\\s=\dfrac{11.1+11.1+20.12}{2}=21.16\\\\\\A=\sqrt{s(s-a)(s-b)(s-c)}\\\\.\ =\sqrt{21.16(21.16-11.1)(21.16-11.1)(21.16-20.12)}\\\\.\ =\sqrt{2227}\\\\.\ =47.1$

Area of shaded region = Area of (1) + Area of (2)

= 247.3 + 47.1

= 294.4

8 0

3 years ago

Read 2 more answers

quezlet: A survey of third- to twelfth-grade students with a sample size of 2026 found that they devoted an average of 1 hour

Maurinko [17]

Answer:

Population is all students from grades 3-12.

Sample is the students who participated in the survey.

Step-by-step explanation:

Given : A survey of third- to twelfth-grade students with a sample size of 2026 found that they devoted an average of 1 hour and 46 minutes per day to using entertainment media.

To find : Identify the population and the sample.

Solution :

The population is defined as it contain all individuals which we want to collect the information.

So, the population consists of all third- to twelfth grade students.

Therefore, Population is all students from grades 3-12.

The sample is defined as a part of population of which information was actually collected.

So, the sample is only the respondents in the survey conducted in 2026.

Therefore, Sample is the students who participated in the survey.

6 0

3 years ago

How to simplify : √(1+√(1-x^2 )) - √(1-√(1-x^2 ))

muminat

√(1 + √(1 - x²)) - √(1 - √(1 - x²))
√(1 + 1 - x) - √(1 - 1 - x)
√(2 - x) - √(0 - x)
(1.414 - x^1/2) - (0 - x^1/2)
1.414 - 0 - x^1/2 + x^1/2
1.414

4 0

3 years ago