¿Cuál es el área de la figura de abajo? 10 cm por 4cm

if a lake has high alkalinity, what is closest to the probability that the lake also has a shallow depth?

never [62]

Answer:

0.22

Step-by-step explanation:

Probability distribution is the function which describes the likelihood of possible values assuming a random variable. The alkalinity of lake is determined by dividing the high shallow depthness by the total of lake alkalinity. The shallow depth is 209 and the total alkalinity of the lake is 966. By dividing the depthness with alkalinity we get 0.22.

209/966 = 0.219

approximately 0.22

8 0

4 years ago

Math can someone please help me I am struggling and about to give up plaese help.

vagabundo [1.1K]

Yes what do you need help with?

4 0

3 years ago

If Linda scored 83 points, what is Lisa's score?

Zepler [3.9K]

Is it a multiple choice question

8 0

4 years ago

Find 1+ 2+ 3+ 4+ ... +50, the sum of the first 50 natural numbers. The sum is

SOVA2 [1]

Step-by-step explanation:

First term(a) = 1

Common difference(d)=2-1=1

No.of term(n)=50

Sn=n÷2(n+1)=50÷2(50+1)=1275

7 0

3 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago