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2 years ago

Please sum 1 help mee

Mathematics

2 answers:

Nimfa-mama [501]2 years ago

5 0

Answer:

the cylinder

Step-by-step explanation:

it is easier to use a calculator to calculate these.

the cone is 56.55 and the cylinder is 62.83

GarryVolchara [31]2 years ago

4 0

Answer:

Step-by-step explanation:

the green cylinder is 20 $\pi$

while the purple cone is 18 $\pi$

which do you think is more ? :)

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2 years ago

WILL GIVE BRAINLIEST& 15 PTS.

kicyunya [14]

Answer:

The given fraction $\frac{x^3-x^2}{x^3}$ reduces to $\frac{x-1}{x}$

Step-by-step explanation:

Consider the given fraction $\frac{x^3-x^2}{x^3}$

We have to reduce the fraction to the lowest terms.

Consider numerator $x^3-x^2$

We can take x² common from both the term,

Thus, numerator can be written as $x^2(x-1)$

Given expression can be rewritten as ,

$\frac{x^3-x^2}{x^3}=\frac{x^2(x-1)}{x^3}$

We can now cancel $x^2$ from both numerator and denominator,

$\Rightarrow \frac{x^2(x-1)}{x^3}=\frac{x^2(x-1)}{x^2 \cdot x}$

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Thus, the given fraction $\frac{x^3-x^2}{x^3}$ reduces to $\frac{x-1}{x}$

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2 years ago

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I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

2 years ago

Please sum 1 help mee​

Please sum 1 help mee