All categories

3 years ago

A rectangle's area is given by the expression 6n2-n+

Mathematics

1 answer:

Zolol [24]3 years ago

8 0

Answer:

Length of rectangle: 2n - (1/3) + 9/(3n)

Width of rectangle = 3n

Rectangle area = A = 6 n^2 - n + 9

Step-by-step explanation:

Rectangle area = A = 6 n^2 - n + 9

test what the roots are: b^2 - 4ac = (-1)^2 - 4*6*9 < 0 no real roots

-215 < 0 discriminant

(6 n^2 - n + 9) / (3n) =

2n - (1/3) + 9/(3n)

If the area is divided by 3n, the quotient is 2n - (1/3) and the remainder is 9

You might be interested in

The polygons in each pair are similar. find the scale factor of the smaller figure to the larger figure.

zaharov [31]

Use the lengths of corresponding sides
so answer is 16:40 which simplifies to 2:5.

4 0

3 years ago

Assume the average height for American women is 64 inches with a standard deviation of 2 inches. A sorority on campus wonders if

satela [25.4K]

Answer:

a) s = 0.4

b) Z = 2.5

c) 99th percentile.

d) The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Assume the average height for American women is 64 inches with a standard deviation of 2 inches

This means that $\mu = 64, \sigma = 2$

A) Calculate the standard error for the distribution of means.

Sample of 25 means that $n = 25$ , so $s = \frac{2}{\sqrt{25}} = 0.4$ .

B) Calculate the z statistic for the sorority group.

Sample mean of 65 means that $X = 65$ .

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{0.4}$

$Z = \frac{65 - 64}{0.4}$

$Z = 2.5$

C) What is the approximate percentile for this sample? Enter as a whole number.

Z = 2.5 has a p-value of 0.9938, so 0.99*100 = 99th percentile.

D) If the sorority actually had 36 members (still with an average of 65 inches), would you expect the percentile value to increase or decrease? why?

The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.

3 0

3 years ago

Can you help me with this question?

Rufina [12.5K]

Answer:

16.2

Step-by-step explanation:

The angle internal to the triangle at B is the supplement of the one shown, so is 65°. That is equal to the angle internal to the triangle at D. Since the vertical angles at C are congruent, the two triangles are similar by the AA theorem.

Corresponding sides of similar triangles are proportional, so we can write the proportion shown in the attachment:

BC/FC = DC/AC

BC = FC(DC/AC) = 21.6(7.2/9.6)

BC = 16.2 . . . . matches the first choice

4 0

3 years ago

the line containing the altitude to the hypotenuse of a right triangle whose vertices are p(-1, 1), q(3, 5), and r(5, -5).

finlep [7]

The equation of the straight line is given by 5y-x=6

An equation can be used to describe a straight line drawn on the Cartesian Plane. These equations have a generic structure and can change based on the slope and where the line intersects the axes.

We will greatly simplify this issue by utilizing the fact that this triangle is specified as a right triangle. We would have had to establish the presence of a right angle if it had been presented as a general triangle.

The segment QR is the hypotenuse of the triangle.

Any side's altitude is located along a line that is perpendicular to that side. We are aware that the slopes of perpendicular lines are negative reciprocals, or:

Slope of line joining Q(3, 5) and R(5, -5)

m= $\frac{y_2-y_1}{x_2-x_1}$