Ask question

Ask question

All categories

Black_prince [1.1K]

3 years ago

13

Mr. Epps buys 500 bags of balloons for the high school prom for $250 what is the cost per bag

2 answers:

Vedmedyk [2.9K]3 years ago

7 0

Hey there,

Lets first look over what info we are given.

500 bags of balloons

$250 for all the bags

To solve this we need to divide the amount of bags of balloons by the price for all of the balloons to find the price per bag.

500 divided by 250 is 2.

So with that we can conclude that $2 dollars is the cost per bag of balloons.

I do hope this helps!
Brainliest is always appreciated if you feel its deserved!

Basile [38]3 years ago

5 0

The answer would be 2

You might be interested in

Find the experimental probability of randomly selecting a marble that is either green or purple. Write your answer in simplest f

EastWind [94]

Green or purple = 10 prob(green or purple) = 10/30 = 1/3

3 0

3 years ago

The $1 coin depicts Sacagawea and her infant son. The diameter of the coin is 26.5 mm, and the thickness is 2.00 mm. Find the vo

Contact [7]

Answer

Find the volume of the coin is cubic millimeters.

To prove

Formula

$Volume\ of\ cylinder\ = \pi r^{2} h$

Where r is the radius and h is the height .

As given

The $1 coin depicts Sacagawea and her infant son.

The diameter of the coin is 26.5 mm, and the thickness is 2.00 mm.

$Radius = \frac{Diameter}{2}$

$Radius = \frac{26.5}{2}$

Radius = 13.25 mm

$\pi = 3.14$

Put in the formula

Volume of coin = 3.14 × 13.25 × 13.25 × 2.00

= 1102.53 mm³ (approx)

Therefore the volume of the coin is 1102.53 mm³ .

7 0

4 years ago

Geometry math question no Guessing and Please show work

galina1969 [7]

here ZLS defines a line segment with end points Z and S.

so it represents line segment ZS

So option B is correct

4 0

3 years ago

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6

USPshnik [31]

LORAN follows an hyperbolic path.

The equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

The coordinates are given as:

$\mathbf{(x,y) = (-60,0)\ (60,0)}$

The center of the hyperbola is

$\mathbf{(h,k) = (0,0)}$

The distance from the center to the focal points is given as:

$\mathbf{c = 60}$

Square both sides

$\mathbf{c^2 = 3600}$

The distance from the receiver to the transmitters is given as:

$\mathbf{2a = 100}$

Divide both sides by 2

$\mathbf{a = 50}$

Square both sides

$\mathbf{a^2 = 2500}$

We have:

$\mathbf{b^2 = c^2 - a^2}$

This gives

$\mathbf{b^2 = 3600 - 2500}$

$\mathbf{b^2 = 1100}$

The equation of an hyperbola is:

$\mathbf{\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1}$

So, we have:

$\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}$

$\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

Hence, the equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

Read more about hyperbolas at:

brainly.com/question/15697124

7 0

3 years ago

Read 2 more answers

A train leaves at 08.43 and arrives at its destination at 09.23. If the train travelled 73 km, what was it's average speed in km

serg [7]

Answer: 109.5 km/ hr

Step-by-step explanation:

Distance = 73 km

Time = 40 minutes = 40/60 = 2/3 hours

Speed = Distance / time

= 73 / 2/3

= 73 x 3/2 = 219 / 2 = 109.5 km/hr

5 0

3 years ago