Ask question

Ask question

All categories

2 years ago

7

You are the owner of the biggest and newest roller coaster called the Gentle Giant. The roller coaster costs $6 to ride. The ope

rator of the ride must pay $200 per day for the ride rental and $65 per day for a safety inspection. If you want to make a profit of at least $1,000 each day, what is the minimum number of people that must ride the roller coaster? Write an inequality that can be used to find the minimum number of people, , which must ride the roller coaster each day to make the daily profit.

1 answer:

viktelen [127]2 years ago

6 0

Answer:

Inequality: 6r - 200 - 65 $\geq$ 1000 and minimum of 211 people

Step-by-step explanation:

Let r = rides

6r - 200 - 65 $\geq$ 1000

6r - 265 $\geq$ 1000

6r $\geq$ 1265

r $\geq$ 210 $\frac{5}{6}$

Needs 211 people to make a profit of at least $1000

You might be interested in

What is 5/6 plus 2 1/8. Thank you

Alecsey [184]

2 47/56
hope this helped

3 0

2 years ago

Read 2 more answers

PLEASEEEEEE HELP!!<br> LOOK AT THE PICTURE PLS!

Serga [27]

The answer is -6 because the square root of 36 is 6, but then you have to add the negative in front of it.

3 0

3 years ago

Read 2 more answers

Which would the phrase "two times the quotient of a number and 3" look like as a variable expression?

Naya [18.7K]

2 n/3 would be the closest answer although it looks as if the 2 is supposed to be multiplied by n/3 which would lead me to write it as 2(n/3)

6 0

3 years ago

EASY PLS HELP! 6= -2(7-c)

Vaselesa [24]

The answer is c = 10

3 0

3 years ago

Read 2 more answers

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago