The sum of the polynomials 6x3 8x2 – 2x 4 and 10x3 x2 11x 9 is

What is the perimeter of this rectangle?

givi [52]

The perimeter is defined as the sum of all the sides of any polygon.

A rectangle is a polygon with 4 sides and 2 pairs of parallel sides. Therefore, in order to find the perimeter of any rectangle, we can either add up all the sides given to you or use a formula which states that:

P = 2L + 2W
Where,
L is the length of the rectangle and W is the width of a triangle.

For example:
A triangle with length = 5 cm and Width = 3 cm.
The Perimeter will be:
P = 2×5 + 2×3
P = 10 + 6
P = 16 cm

3 0

3 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago

Which value can be added to the set below without changing its mean?

shusha [124]

Answer:

15..

Step-by-step explanation:

4, 10, 14, 18, 22, 22

Mean = (4 + 10 + 14 + 18 + 22 + 22)/6

= 90/6

= 15.

Number to be added, 'y'

For the mean to be the same

(90 + y) / 7 = 15

90 + y = 15 × 7

90 + y = 105

y = 105 - 90

= 15

7 0

3 years ago

2 + 1/6a = -4 what is a

horsena [70]

Answer:

I found that it was -1, but everything online is telling me -36.

6 0

3 years ago

Factor 4x^4yz-16y^3z

Iteru [2.4K]

Answer:

Step-by-step explanation:

$4x^4yz-16y^3z\\\\4x^4yz=4yz\cdot x^4\\\\16y^3z=4yz\cdot4y^2\\\\4x^4yz-16y^3z=4yz\cdot x^4-4yz\cdot4y^2=4yz(x^4-4y^2)\\\\x^4-4y^2=x^{2\cdot2}-2^2y^2=(x^2)^2-(2y)^2=(x^2-2y)(x^2+2y)\\\\Used:\\\\(a^n)^m=a^{nm}\\\\(ab)^n=a^nb^n\\\\a^2-b^2=(a-b)(a+b)\\\\4x^4yz-16y^3z=4yz(x^2-2y)(x^2+2y)$

6 0

3 years ago