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3 years ago

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hurry! You receive a $150 bonus each month you reach your sales quota. You reach your sales quota for each of 3 months. What is

the total amount you receive in bonuses for the three months? $50 $300 $350 $450 $600

1 answer:

Lera25 [3.4K]3 years ago

7 0

450

I used my head lol

You might be interested in

200 σ j=1 2j( j 3) describe the steps to evaluate the summation. what is the sum?

ziro4ka [17]

The sum of the equation is = 5494000.

<h3>What does summation mean in math?</h3>

The outcome of adding numbers or quantities mathematically is a summation, often known as a sum. A summation always has an even number of terms in it. There may be just two terms, or there may be 100, 1000, or even a million. Some summations include an infinite number of terms.

<h3>Briefing:</h3>

Distribute 2j to (j+3).

Rewrite the summation as the sum of two individual summations.

Evaluate each summation using properties or formulas from the lesson.

The lower index is 1, so any properties can be used.

The sum is 5,494,000.

<h3>Calculation according to the statement:</h3>

$\sum_{j=1}^{200} 2 j(j+3)$

simplifying them we get:

$\sum_{j=1}^{200} 2 j^{2}+6 j$

Split the summation into smaller summations that fit the summation rules.

$\sum_{j=1}^{200} 2 j^{2}+6 j=2 \sum_{j=1}^{200} j^{2}+6 \sum_{j=1}^{200} j$

$\text { Evaluate } 2 \sum_{j=1}^{200} j^{2}$

The formula for the summation of a polynomial with degree 2

is:

$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$

Substitute the values into the formula and make sure to multiply by the front term.

$(2)$$\left(\frac{200(200+1)(2 \cdot 200+1)}{6}\right)$$$

we get: 5373400

Evaluating same as above : $6 \sum_{j=1}^{200} j$

we get: 120600

Add the results of the summations.

5373400 + 120600

= 5494000

The sum of the equation is = 5494000.

To know more about summations visit:

brainly.com/question/16679150

#SPJ4

6 0

2 years ago

PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

----------------------------------------

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

--------------------------

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

7 0

3 years ago

Read 2 more answers

What is 25 divided by 625?

scoray [572]

25/ 625 = 5 / 125 = 1/15

8 0

3 years ago

Rewrite the function by completing the square <br> f(x)=x^2-14x+63

Veseljchak [2.6K]

Answer:

f(x) = (x - 7)² - 14

General Formulas and Concepts:

<u>Pre-Algebra</u>

Equality Properties

<u>Algebra I</u>

Standard Form: ax² + bx + c = 0
Vertex Form: f(x) = a(bx - c)² + d
Completing the Square: (b/2)²

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = x² - 14x + 63

<u>Step 2: Rewrite</u>

Separate: f(x) = (x² - 14x) + 63
Complete the Square: f(x) = (x² - 14x + 49) + 63 - 49
Simplify: f(x) = (x - 7)² - 14

8 0

3 years ago

Read 2 more answers

You are given the information that P(A) = 0.30 and P(B) = 0.40.

Ad libitum [116K]

Answer:

1.B. No. You need to know the value of P(A and B). 2.C. Yes P(A and B) =0, so P(A or B) = P(A) + P(B).

Step-by-step explanation:

We can solve this question considering the following:

For two mutually exclusive events:

$\\ A_{1}\;and\;A_{2}$

$\\ P(A_{1} or A_{2}) = P(A_{1}) + P(A_{2})$ (1)

An extension of the former expression is:

$\\ P(A_{1} or A_{2}) = P(A_{1}) + P(A_{2}) - P(A_{1} and A_{2})$ (2)

In <em>mutually exclusive events,</em> P(A and B) = 0, that is, the events are <em>independent </em>one of the other, and we know the probability that <em>both events happen</em> <em>at the same time is zero</em> (P(A <em>and</em> B) = 0). There are some other cases in which if event A happens, event B too, so they are not mutually exclusive because P(A <em>and</em> B) is some number different from zero. Notice the difference between <em>OR</em> and <em>AND. The latter implies that both events happen at the same time.</em>

In other words, notice that the formula (2) provides an extension of formula (1) for those events that are not <em>mutually exclusive</em>, that is, there are some cases in which the events share the same probabilities in a way that these probabilities <em>must be subtracted</em> from the total, so those probabilities in common do not "inflate" the actual probability.

For instance, imagine a person going to a gas station and ask for checking both a tire and lube oil of his/her car. The probability for checking a tire is P(A)=0.16, for checking lube oil is P(B)=0.30, and for both P(A and B) = 0.07.

The number 0.07 represents the probability that <em>both events occur at the same time</em>, so the probability that this person ask for checking a tire or the lube oil of his/her car is:

P(A or B) = 0.16 + 0.30 - 0.07 = 0.39.

That is why we cannot simply add some given probabilities <em>without acknowledging if the events are or not mutually exclusive</em>, whereas we can certainly add the probabilities in question when we know that both probabilities are <em>mutually exclusive</em> since P(A and B) = 0.

In conclusion, knowing the events are mutually exclusive <em>does</em> provide <em>extra information</em> and we can proceed to simply add the probabilities of either event; thus, the answers are those in which <em>we need to previously know the value of P(A and B)</em>.

7 0

3 years ago