The human resources manager at a company records the length in hours, of one shift at work, X. He creates the probabity

D is not right its a negative 8 so therefore it would be minus

Ilya [14]

I really dont understant your question. but you are right that negative 8 that it would be minus. In mathematics, a negative number is a real number that is less than zero. Negative numbers represent opposites. If positive represents movement to the right, negative represents movement to the left.

6 0

3 years ago

Which points are separated by distance of 6 units?

sergiy2304 [10]

I think it’s b! i hope that helps!

8 0

2 years ago

A sheep is bought for £40 and sold for £33. What is the percentage loss?

valkas [14]

Answer:

£13 %11

Step-by-step explanation:

4 0

2 years ago

On a trip to Florida, Santina drove 203 miles and Carole drove 189 miles. How many more miles did Santina drive than Carole?

matrenka [14]

Answer:

1. santina drove 14 more miles than carole

2. the owner will need to buy 8 bags this week

Step-by-step explanation:

1. 203-189= 14

2. if they use 80 pounds a week and can only buy 10 pounds in one bag, they'll need 8 bags to equal 80 pounds since each bag is 10 pounds

3 0

1 year ago

You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was

Doss [256]

Answer:

A) The linear relation between price and demand is:

$d=-550x+2750$

The revenue R is:

$R=-550x^2+2750x$

B) The profit functionP is:

$P=-550x^2+2750x-30$

C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.

Step-by-step explanation:

We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.

We have two points for this linear relationship:

At price x=3, the demand is d=1100.
At price x=2.5, the demand is d=1375.

We will model the relation:

$d=mx+b$

We can calculate the slope m as:

$m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550$

Then, replacing one point in the linear equation, we can calculate the intercept b:

$d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750$

Then, the linear relation between demand and price is:

$d=-550x+2750$

The revenue R can be expressed as the multiplication of the price and the demand:

$R=x\cdot d=x(-550x+2750)=-550x^2+2750x$

If we have a fixed cost of $30 per month, the profit P is:

$P=R-FC=-550x^2+2750x-30$

We can maximize the profit by deriving the profit function and making it equal to zero.

$\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5$

This corresponds to a profit of:

$P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5$

5 0

3 years ago