A rectangle has a height of 5m^4 and a width of m^2-2m-1.

Mathematics

2 answers:

In-s [12.5K]2 years ago

8 0

Answer:

area of the rectangle = 5(m⁶ - 2m⁵ - m⁴) or 5m⁴(m² - 2m - 1)

Step-by-step explanation:

The rectangle has a height of 5m⁴ and the width is m² - 2m - 1.The area of the rectangle is the product of the width and the height.

width = m² - 2m - 1

height = 5m⁴

area of rectangle = 5m⁴ × (m² - 2m - 1)

area of rectangle = 5m⁴(m² - 2m - 1)

area of the rectangle = 5m⁶ - 10m⁵ - 5m⁴

area of the rectangle = 5(m⁶ - 2m⁵ - m⁴)

The area of the entire rectangle can be expressed as 5(m⁶ - 2m⁵ - m⁴) or 5m⁴(m² - 2m - 1)

Serhud [2]2 years ago

8 0

Answer:

5m^6-10m^5-5m^4

Step-by-step explanation:

You might be interested in

What is 56/19 as a mixed number

pashok25 [27]

The answer is 2 18/19

6 0

2 years ago

Read 2 more answers

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: