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3 years ago

Hello can you please help me posted picture of question

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

3 0

Since the order of selection does not matter we will use combinations to solve this problem.

Total number of students = 20
Number of students to be selected = 4

We are to find the combinations of 20 objects taken 4 at a time. This can be expressed as 20C4

20C4 = 4845

Thus, there are 4845 ways to select the 4 students.

So, the correct option is D

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Find the value of Z please help

JulijaS [17]

I think it is 1 because it is smaller than 1.5 so it would have to be 1

4 0

3 years ago

Sharon was going through the financial records of her company. The profit earned by the company, p(t), over time t, in years, fo

FromTheMoon [43]

For this case what we must do is find a quadratic function that is already factored.
This is because in the factored quadratic equations, it is easier to observe the zeros of the function.
In this case, the zeros of the function represent the time at which the company did not make any profit.
We have the following equation:
p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
We observed that there was no gain in:
t = 3
t = 5
The other roots are discarded because they are negative
Answer:
a.p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)

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3 years ago

What digit makes each number divisible by nine?<br><br> _7,302

Alborosie

Answer:

Not completely sure but hopefully these will help.

Step-by-step explanation:

I'm kind of confused what you're asking for. If you could clarify that would be great. If you're trying to make it divisible by 9, you can either do 3, 6 or 9. If you want the first two digits: 2,5,8. Please clarify so I can help

8 0

3 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

3 years ago

Round 2826 to the nearest hundred. PLEASE HELP ME ASAP!!!

kotykmax [81]

It is 30000 i think dang it

5 0

3 years ago

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