Find the least common denominator between the two fractions. 1/2 and 5/6 2 6 8 12

Find the slope of the tangent line to the graph of the function at the given point.

rodikova [14]

Well, we see that it is a straight line
so the derivitive would be perpendicular
5/4 times what=-1
-4/5

the slope is -4/5

7 0

3 years ago

The sum of a number and its square is 182. Find the number.

Mars2501 [29]

Answer:

13 and -14 satisfy this condition

Step-by-step explanation:

Let's represent that number as x

and the square of x is x^2

So,

x + x^2 = 182

Subtract 182 from both sides

x + x^2 - 182 = 182 - 182

x + x^2 - 182 = 0

rearrange the quadratic equation

x^2 + x -182 = 0

let's use the quadratic formula

$\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

a = 1, b = 1, c = -182

$\frac{-1+\sqrt{1^2-4*1*(-182)} }{2*1}$ or $\frac{-1-\sqrt{1^2-4*1*(-182)} }{2*1}$

$\frac{-1+\sqrt{1+728} }{2}$ or $\frac{-1-\sqrt{1+728} }{2}$

$\frac{-1+\sqrt{729} }{2}$ or $\frac{-1-\sqrt{729} }{2}$

$\frac{-1+{27} }{2}$ or $\frac{-1-{27} }{2}$

$\frac{26}{2}$ or $\frac{-28}{2}$

13 or - 14

Lets check

13 + 13^2 = 13 + 169

= 182

Also,

-14 + (-14^2) = -14 + 196

= 182

8 0

3 years ago

Confidence Interval Mistakes and Misunderstandingsâ€”Suppose that 500 randomly selected recent graduates of a university were as

kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

$n=500\\\bar x=8.6\\\sigma=2.2$

(a)

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

$8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)$

(b)

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

n = 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0

3 years ago

Which of the following diagram represent a relation that is not a function

andrew11 [14]

You didn’t even post a diagram

which of the following diagram represent a relation that is not a function

8 0

3 years ago

Jon's weight loss for each week of the month is 4lbs., 3.5 lbs. and 2 lbs. He gained 4.5 lbs. the last week. If Jon originally w

aalyn [17]

Since he gained 4.5 lbs but he lost 4 + 3.5 + 2 = 9.5 lbs, in total, he lost 5 lbs.
hope it helps!
hit the thank you button!

4 0

3 years ago