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jeyben [28]
3 years ago
11

As a bowling instructor, you calculate your students' averages during tournaments. In 5 games, one bowler had the following scor

es: 143, 156, 172, 133, and 167. What was that bowler's average?
Mathematics
1 answer:
erastovalidia [21]3 years ago
4 0

Answer:

154.2

Step-by-step explanation:

143 plus

156 plus

172 plus

133 plus

167 = 771

divide by 5 equals 154.2

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Answer:

1.) 7

2.)7

3.)-7

4.)7

5.)7

Step-by-step explanation:

1.) Just count -2 to 5 with your hands

2.) since two negative signs make a addition sign when they are close to each other, you will just add 5 and 2.

3.) Just add 5 and 2 and add the negative sign before the number.

4.)Same as I said in number 2, it stays 7.

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Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x
soldier1979 [14.2K]

Answer:

(a) The function is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is x = -\frac{1}{2}

(d) The function is concave upward on \left(- \frac{1}{2}, \infty\right) and concave downward on \left(-\infty, - \frac{1}{2}\right)

Step-by-step explanation:

(a) To find the intervals where f(x) = 2x^3 + 3x^2 -180x is increasing or decreasing you must:

1. Differentiate the function

\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6

These points divide the number line into three intervals:

(-\infty,-6), (-6,5), and (5, \infty)

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right

Therefore f(x) is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

  • f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.
  • f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6

We set f''(x) = 0

f''(x) =12x+6 =0\\\\x=-\frac{1}{2}

Analyzing concavity, we get

\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right

The function is concave upward on (-1/2,\infty) because the f''(x) > 0 and concave downward on (-\infty,-1/2) because the f''(x) < 0.

f(x) is concave down before x = -\frac{1}{2}, concave up after it. So f(x) has an inflection point at x = -\frac{1}{2}.

7 0
3 years ago
a cell phone company test 95 cell phones and finds that seven of them have defects out of the 760 cell phones how many would you
suter [353]

Answer:

56

Step-by-step explanation:

simple ratio: 7/95 = x/760

first, calculate 7/95 which is 0.0736

next, multiply that by 760, which would be 56

6 0
3 years ago
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