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2 years ago

Pyramid = 40

Mathematics

1 answer:

Y_Kistochka [10]2 years ago

5 0

Answer:

A. 39,000.

Step-by-step explanation:

Volume of the lower prism = 30^3= 27,000

Volume of the pyramid = 1/3 area of the base * height

= 1/3 * 900 * 40

= 1200

Volume of the whole solid = 27000 + 12000

= 39,000,

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Two sections of a class took the same quiz. Section A had 15 students who had a mean score of 80, and Section B had 20 students

Nezavi [6.7K]

Answer: A. 85.7

Step-by-step explanation:

Given : Two sections of a class took the same quiz.

Section A had 15 students who had a mean score of 80, and Section B had 20 students who had a mean score of 90.

We know that , $\text{Mean}=\dfrac{\text{Sum of observations}}{\text{No. of observations}}$

Then , for section A :

$\text{Mean score }=\dfrac{\text{Sum of scores in sec A}}{\text{No. of students}}$

$\Rightarrow\ 80=\dfrac{\text{Sum of scores in sec A}}{15}\\\\\Rightarrow\ \text{Sum of scores in sec A}=80\times15=1200$

Similarly in Section B, $\text{Sum of scores in sec B}=90\times20=1800$

Total scores = Sum of scores in sec A+Sum of scores in sec B

=1200+1800=3000

Total students = Students in sec A +Students in sec B

=15+20=35

Now , the mean score for all of the students on the quiz = $\dfrac{\text{Total score}}{\text{Total students}}$

$=\dfrac{3000}{35}=85.7142857143\approx85.7$

Hence, the approximate mean score for all of the students on the quiz = 85.7

Thus , the correct answer is option A. 85.7.

7 0

2 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$