If gumballs cost 150 a bag ,and there 15!gumballs in each bag much do you need to charge for each gumball in order to make money

2 answers:

7 0

More than 10 a gumball.

150 divided by 15 is 10, which would be the price you bought for. If you want to make more money, you must at least have a surge pricing of 10 unit prices per gumball.

Natalka [10]2 years ago

5 0

10 it is 10 because if u divide 150 and 15 you would get 10

You might be interested in

HELP!!!!!!!!!!!!!!!!!!!!<br><br> what is 3x4^2<br><br> EXPLAINN

uranmaximum [27]

Answer:

48x

PEMDAS: Do the exponent first 4 times 4 is 16 times that by 3x to get 48x.

3 0

2 years ago

Which expressions are equivalent to 24 + 8? Choose ALL that apply,

zepelin [54]

Answer:
4(6+2)
2(12+4)
8(3+1)
Hope that helps :)

4 0

2 years ago

Read 2 more answers

PLEASE HELP ASAP!! the slope of a line is -2. Two points on the line are (eight, a) and (2, 4). what is the value of a?

Serggg [28]

Answer:

a=4

Step-by-step explanation:

5 0

2 years ago

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

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

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

2 years ago

The phony bologna meat company offers two platters. platter a comes with 4 hamburgers, 3 hot dogs, and 2 pig's feet, and costs $

Brrunno [24]

The solution for this problem is:
If there is 60 platters of B at a cost of $720:
(220 - 60 x 3) / 4 = 10 platters of A to make up for the deficit in hamburgers
(270 - 60 x 4) / 3 = 10 platters of A to make up for the deficit in hot dogs
(250 - 60 x 5) / 2 = 0 platters of A since there is no deficit in pigs feet

So 10 platters A are required at a cost of $150. $720 + $150 = for a total minimum cost of $870.

8 0

3 years ago

Read 2 more answers