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3 years ago

Experimental/theoretical

Mathematics

2 answers:

Lera25 [3.4K]3 years ago

8 0

Answer:

A.200

Step-by-step explanation:

ale4655 [162]3 years ago

3 0

Answer:

200

Step-by-step explanation:

Ella's name appears on the table 2/10 times, so we can multiply the denominator to make it 1000.

10x100 = 1000

If we multiply the denominator by 100, we must do the same to the numerator,

2x100=200

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Given: 315/1575. Write the prime fractorization of the numerator and the denominator, then write the fraction in lowest terms.

Ne4ueva [31]

The prime factorization of the numerator, 315, is 5 x 3 x 3 x 7. That of the denominator, 1575 is 5 x 5 x 3 x 3 x 7. To convert the fraction to its lowest form, cancel all the factors that appear to both the numerator and denominator. We will be left with 1/5.

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3 years ago

Find the median of the data.

kipiarov [429]

Answer:

92.5

Step-by-step explanation:

The data given:

93, 81, 94, 71, 89, 92, 94, 99

Put the data in the ascending order:

71, 81, 89, 92, 93, 94, 94, 99

Since the data size is even, the median is the average of middle two:

median = (92 + 93)/2 = 92.5

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2 years ago

You invest $1,050.00, buying stock in a company at $10.00/share. The shares increase in value by 13% the first year and decrease

allochka39001 [22]

Number of shares
1050/10=105
After one year share value
10×(1+0.13)=11.3
After the second year share value
11.3×(1−0.05)=10.735
Total
10.735×105=1,127.175

3 0

3 years ago

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Elanso [62]

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3 years ago

Read 2 more answers

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago