Ask question

Ask question

All categories

3 years ago

7

The Anderson family is preparing for a family reunion. They think they will need 22 juice boxes for every 6 children who attend.

How many juice boxes will they need if they are expecting 12 children?
44 juice boxes

22 juice boxes

264 juice boxes

132 juice boxes

1 answer:

AleksandrR [38]3 years ago

6 0

If they need 22 boxes for every 6 kids the they would need 44 boxes for every 12 kid

You might be interested in

Ok please help meeeeeeeeeeeeeeeee

valina [46]

Answer:

Step-by-step explanation:

Volume of a cone is given by the formula,

V = $\frac{1}{3}\pi r^{2}h$

Here, r = Radius of the circular base of the cone

h = Height

By substituting values from the question,

V = $\frac{1}{3}\times \pi (\frac{6}{2})^2(3)$

V = 9π

V = 28.27 m³ [Nearest hundred]

If we have to round the answer to the nearest tenth, answer will be,

Volume = 28.3 m³

If we have have to round the answer to the whole number then the answer will be,

Volume = 28 m³

8 0

3 years ago

What is 1 dollor 2 dimes 13 pennies PLUS 2 dollaars and 3 quqrters?

gayaneshka [121]

$4.08

Explanation:
That would be a lot of addition, and $4 and 8 cents.
Since $1+$2= $3
2 d + 3 q = 95c
95c + 13c = $1.08
$1.08 + $3 = $4.08

8 0

2 years ago

Read 2 more answers

Explain what the slope and intercept mean in the situation.

Strike441 [17]

Answer:

m=3/1 y=3x+0

Step-by-step explanation:

as the perimeter goes up 3 the side lengths go up one.

hope this helps

is this the 3.6 more linear relationships worksheet?

cause if so i have the answer sheet lol

6 0

3 years ago

What is 5.316 - 1.942 (show ur work)

Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

$=3.374$

Hence the correct answer is 3.374

7 0

2 years ago

How do you find the phase shift of the trig function y=4sin(pieX+2)-5

max2010maxim [7]

The phase shift is the value of the angle when the variable
in the angle is zero.

The angle in the sin function in this equation is (pi X + 2).

When the variable 'X' is zero, the angle is 2 .

So the phase shift of this sin function is 2 (of whatever
the unit of the angle is ... 2 degrees, 2 radians, etc.)

5 0

3 years ago