What is x2 – 6 = 16x + 30 rewritten in standard form, then factored?

What decimal best represents an elevation of 2,000.288 feet below sea level

choli [55]

I believe the answer is -2,000.288 feet but im not 100% sure. hope this helps, have an amazing day :)

3 0

4 years ago

15 POINTS!

Oliga [24]

Answer:

y = 2x + 1

Step-by-step explanation:

Slope-int form: y = mx + b
m = slope or rise/run
b = y-intercept

Given:

Point (2, 5)
m (slope) = undefined

When a slope is undefined, the line is a vertical line (it rises but doesn't "run").

To find m, use the given point's x-coordinate because the line you're trying to find will run vertically through that point.

(x, y) = (2, 5); x = 2
m = 2

Now the equation becomes y = 2x + b.

To find b, substitute the given point and the found slope into the formula, then solve for b.

5 = 2(2) + b
5 = 4 + b
5 - 4 = b
b = 1

Now that you know what b is, you can substitute b into the equation:

y = 2x + 1

8 0

3 years ago

6 = m/8 <br> M=?<br> What is the answer for M?

galben [10]

Answer:

m =48

Step-by-step explanation:

1st step is to multiply both sides by 8

2nd step it would be 48=m

3rd step swap the sides.. m=48

6 0

3 years ago

A researcher is concerned about the impact of students working while they are enrolled in classes, and she likes to know if stud

8_murik_8 [283]

Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(c) Margin of error = 0.800

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

$\bar x = 7.10$ -- sample mean

$\sigma=5$ --- sample standard deviation

$n = 150$ --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

$CI = 90\%$ --- the confidence interval

Calculate the $\alpha$ level

$\alpha = 1 - CI$

$\alpha = 1 - 90\%$

$\alpha = 1 - 0.90$

$\alpha = 0.10$

Divide by 2

$\frac{\alpha}{2} = 0.10/2$

$\frac{\alpha}{2} = 0.05$

Subtract from 1

$1 - \frac{\alpha}{2} = 1 - 0.05$

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

From the z table. the critical value for $1 - \frac{\alpha}{2} = 0.95$ is:

$z = 1.960$

Solving (c): Margin of error

This is calculated as:

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

$E = 1.960 * \frac{5}{\sqrt {150}}$

$E = 1.960 * \frac{5}{12.25}$

$E = \frac{1.960 *5}{12.25}$

$E = \frac{9.80}{12.25}$

$E = 0.800$

Solving (d): The confidence interval

This is calculated as:

$CI = (\bar x - E, \bar x + E)$

$CI = (7.10 - 0.800, 7.10 + 0.800)$

$CI = (6.3, 7.9)$

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

6 0

3 years ago

Which of the following statements is true?

ra1l [238]

Ur answer is B............!!!!!!

Hope its right!!

7 0

3 years ago

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