The value of -8/15 · 20/64 is _____. 1) -1/6 2) 1/6 3) 2/3 4) -2/3

The cost function for a certain company is C = 20x + 700 and the revenue is given by R = 100x − 0.5x2. Recall that profit is rev

Brilliant_brown [7]

Answer:

x = 20

x = 140

Step-by-step explanation:

Find the equation for profit P

P = R - C

Substitute the equations for R and C then simplify.

P = 100x − 0.5x² - (20x + 700)

P = -0.5x² + 80x - 700

Find the values of x when profit is $700

P = -0.5x² + 80x - 700

700 = -0.5x² + 80x - 700

0 = -0.5x² + 80x - 1400

This is in the standard form 0 = ax² + bx + c

Use the quadratic formula to find values of x

$x = \frac{-b±\sqrt{b^{2}-4ac} }{2a}$

(Ignore Â)

Substitute a b and c.

a = -0.5

b = 80

c = -1400

$x = \frac{-(80)±\sqrt{80^{2}-4(-0.5)(-1400)} }{2(-0.5)}$

$x = \frac{-80±60}{-1}$

Split the formula at the ± so that there are two to get the two x values.

$x = \frac{-80+60}{-1}$

$x = \frac{-80-60}{-1}$

x = 20

x = 140

The profit will be $700 when x is 20 or when x is 140.

5 0

2 years ago

At eagle bay it cost $12 per hour to rent a Canoe Nate and his friends rented a canoe for 4 hours and paid $68. Write and solve

stich3 [128]

The cost to rent the canoe for 7 hours is $ 104

Solution:

At eagle bay it cost $12 per hour to rent a Canoe

Nate and his friends rented a canoe for 4 hours and paid $68

Let "x" be the number of hours rented

We can solve use the point slope equation

$y-y_1 = m(x-x_1)$

Where, "m" is the rate of change

$(x_1, y_1)$ is the sample point

From given,

m = 12

$(x_1, y_1) = (4, 68)$

Therefore,

$y - 68 = 12(x-4)$

$y - 68 = 12x -48\\\\y = 12x -48+68\\\\y = 12x + 20$

Thus the linear equation is found

Here, "y" represents the amount paid for "x" hours

Find the cost to rent the canoe for 7 hours

Substitute x = 7 in linear equation

y = 12(7) + 20

y = 84 + 20

y = 104

Thus cost to rent the canoe for 7 hours is $ 104

4 0

2 years ago

For a certain type of copper wire, it is knownthat, on the average, 1.5 flaws occur per millimeter.Assuming that the number of f

mario62 [17]

Answer:

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters

Step-by-step explanation:

Step 1:-

Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.

by Poisson random variable given that λ = 1.5 flaws/millimeter

Poisson distribution $P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}$

Step 2:-

The probability that no flaws occur in a certain portion of wire

$P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}$

On simplification we get

P(x=0) = 0.223 flaws occur / millimeters

Conclusion:-

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters

5 0

2 years ago

The orbital period, P, of a planet and the planet's distance from the sun, a, in astronomical units is related by the formula P

anygoal [31]

Answer:

a for saturn is 9.55 astronomical units

Step-by-step explanation:

Here, we are interested in calculating the distance from the sun.

Mathematically, from the question;

P = a^3/2

29.5 = a^3/2

a = 29.5^2/3

a = [3√(29.5)]^2

a = 9.55 in astronomical units

3 0

2 years ago

Read 2 more answers

On Sunday Sheldon bought 3 and 1/2kg of plant food. He used 1 and 2/3kg on his strawberry plants and used 1/4kg for his tomato p

notsponge [240]

The remaining plant food is: 1 7/12 kg

Step-by-step explanation:

Given

Total Plant food = 3 1/2 kg

Plant food used on Strawberry Plants = 1 2/3 kg

Plant food used on tomato Plants = 1/4 kg

We will first calculate the total food plant used and then subtract the used plant food from the total plant food.

So,

$Total\ Plant\ Food\ used=1\frac{2}{3}+\frac{1}{4}\\=\frac{5}{3}+\frac{1}{4}\\=\frac{20+3}{12}\\=\frac{23}{12}$

Now,

$Left\ Food\ Plant=3\frac{1}{2}-\frac{23}{12}\\=\frac{7}{2}-\frac{23}{12}\\=\frac{42-23}{12}\\=\frac{19}{12}\\=1\frac{7}{12}$

The remaining plant food is: 1 7/12 kg

Keywords: Division, Fractions

Learn more about fractions at:

#LearnwithBrainly

7 0

2 years ago