Which mathematician is not known as a contributor to projective geometry?

The volume of a box is given by the expression

Kamila [148]

Answer:

Area = x^2+4x+3

Step-by-step explanation:

The volume of a box is given by the expression

x^3 + 3x^2 - x-3. The height of the box is given by the expression x-1.

volume of the box = Length * width * height

L is the side length of the box;

Given

V(x) = x^3 + 3x^2 - x-3

Factorize

V(x) = x^2(x+3)-1(x+3)

V(x) = (x^2-1)(x+3)

V(x) = (x+1)(x-1)(x+3)

V = LWH

V = Area * Width

Since the height of the box is x - 1

Area = (x+1)(x+3)

Expand

Area = x^2+3x+x+3

Area = x^2+4x+3

3 0

3 years ago

Number Height of

svlad2 [7]

Hey there, Selenawebb0418!

If you look at the table, you can see that the plant grows 2 centimeters (I'm assuming it's centimeters) every 3 days.
So, the rate of the plant should be $\frac{2}{3}$ cm per day.

Thank you for using Brainly.
See you soon!

5 0

4 years ago

a zoologist hides in a bush 200 feet away from a tree inhabited by a family of monkeys. if the zoologist is looking up at an ang

Nastasia [14]

Answer:

$\displaystyle 53.6ft$

Step-by-step explanation:

refer the attachment

let the height be h

since we are given the adjacent side and want to figure out opposite side we'll use tan function

so our equation is

$\displaystyle \tan( {15}^{ \circ} ) = \frac{h}{200}$

cross multiplication:

$\displaystyle h = 200\tan( {15}^{ \circ} )$

by using calculator we acquire:

$\displaystyle h \approx 53.6ft$

hence,

about 53.6ft high in the tree ate monkeys

4 0

3 years ago

The ________ could be used to describe a data set by itself, while ___________ would almost never be used to describe a data set

Alona [7]

The mean could be used to describe a data set by itself, while interquartile range would almost never be used to describe a data set by itself.

<h3>What is Descriptive statistics?</h3>

These are operations which are used to summarize certain characteristics a set of data has.

Mean and range deals with the full set of data while interquartile range measures just the middle half of the data and is denoted as option A.

Read more about Descriptive statistics here brainly.com/question/6990681

#SPJ1

8 0

2 years ago

Use the given information to find the p-value. Also, use a 0.05 significance level and state the conclusion about the null hyp

faltersainse [42]

Answer:

$p_v =P(z>1.34)=1-P(z$

Step-by-step explanation:

1) Data given and notation n

n represent the random sample taken

Xrepresent the people with a characterisitc in the sample

$\hat p$ estimated proportion of people with the characteristic desired

$p_o=0.554$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the population proportionis higher than 0.554.:

Null hypothesis: $p\leq 0.554$

Alternative hypothesis: $p > 0.554$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

The value of the statisitc is already calculate and given:

$z=1.34$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a one right tailed test the p value would be:

$p_v =P(z>1.34)=1-P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.554 .

3 0

3 years ago