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3 years ago

Write the equation of a line with a slope of −1 and a y-intercept of −6. (2 points)

Mathematics

1 answer:

tatyana61 [14]3 years ago

3 0

Answer:

y = -1x - 6

Step-by-step explanation:

Hi!

The format of the equation of a line is y = mx + b

m is the slope and b is the y-intercept

We just need to substitute in the values you gave for the variables and we get the answer,

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Sedaia [141]

Answer:

The first graph

Step-by-step explanation:

when x is equal to something it is always up

y = 2 would be #4

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3 years ago

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5 b = - 25 for algebra

san4es73 [151]

Answer:

b= -5

Step-by-step explanation:

because 5x-5=-25

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2 years ago

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What is (8-√2)(2+√8)?

siniylev [52]

Answer:

$(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = 12 + 14\; \sqrt{2}$ .

Step-by-step explanation:

Step One: Simplify the square roots.

$8 = 4 \times 2 = 2^2 \times 2$ .

The square root of 8 $\sqrt{8}$ can be simplified as the product of an integer and the square root of 2:

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{2^2 \times 2} = \sqrt{2^{2}} \times \sqrt{2} = 2 \; \sqrt{2}$ .

As a result,

$(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = (8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})$ .

Step Two: Expand the product of the two binomials.

$(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})$ is the product of two binomials:

Binomial One: $8 - \sqrt{2}$ .
Binomial Two: $2 + 2\; \sqrt{2}$

Start by applying the distributive law to the first binomial. Multiply each term in the first binomial (without brackets) with the second binomial (with brackets)

$({\bf 8} - {\bf \sqrt{2}}) \cdot {(2 + 2\; \sqrt{2})}\\= [{\bf 8} \cdot {(2 + 2\; \sqrt{2})}] - [{\bf \sqrt{2}} \cdot {(2 + 2\; \sqrt{2})}]$

Now, apply the distributive law once again to terms in the second binomial.

$[8 \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})] - [\sqrt{2} \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})]\\= [8 \times {\bf 2} + 8 \times {\bf 2\;\sqrt{2}}] - [\sqrt{2} \times {\bf 2} + \sqrt{2} \times {\bf 2\; \sqrt{2}}]$ .

Step Three: Simplify the expression.

The square of a square root is the same as the number under the square root. For example, $\sqrt{2} \times \sqrt{2} = (\sqrt{2})^{2} = 2$ .

$[8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + 2 \sqrt{2} \times \sqrt{2}]\\ =[16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4$ .

Combine the terms with the square root of two and those without the square root of two:

$16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})$ .

Factor the square root of two out of the second term:

$(16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\= (16 - 4) + (16 - 2) \; \sqrt{2} \\= 12 - 14 \; \sqrt{2}$ .

Combining the steps:

$(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})\\= [8 \cdot (2 + 2\; \sqrt{2})] - [\sqrt{2} \cdot (2 + 2\; \sqrt{2})]\\= [8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + \sqrt{2} \times 2 \;\sqrt{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times (\sqrt{2})^{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times 2]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\$

$= (16 - 4) + (16 - 2) \; \sqrt{2}\\= 12 - 14 \; \sqrt{2}$ .

3 0

3 years ago

First derivative of this function ?

german

Use the chain rule.

$f'(x) = \left[\left(x^2 + \cos^2(x)\right)^3\right]' \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left[x^2 + \cos^2(x)\right]' \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left(\left[x^2\right]' + \left[\cos^2(x)\right]'\right) \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left(2x + 2 \cos(x) \left[\cos(x)\right]'\right) \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left(2x - 2 \cos(x) \sin(x)\right) \\\\ ~~~~ = \boxed{3 \left(x^2 + \cos^2(x)\right)^2 \left(2x - \sin(2x)\right)}$

7 0

2 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

8 0

3 years ago