All categories

3 years ago

Ohio City Burgers charges $6.75 for a

Mathematics

1 answer:

luda_lava [24]3 years ago

3 0

Answer: x = (10 - 6.75) ÷ 0.75

Step-by-step explanation:

x stands for the maximum number of toppings you can afford.

10 stands for the amount of money you have (10$).

6.75 is the amount of money spent on the Burger alone.

0.75 is the money spent on each topping with your remaining money.

You might be interested in

ALMOST 50 POINTS COME AND GET IT!!!

marusya05 [52]

Answer:

yay tysmm also good morning

7 0

3 years ago

Read 2 more answers

Which of the following represents the graph of f(x) = 4X – 2?

alina1380 [7]

Answer:

The bottom one.

Step-by-step explanation:

3 0

3 years ago

PLEASE HELP URGENT

aivan3 [116]

(10/17) / (-15/17) =
10/17 * - 17/15 =
- 10/15 =
- 2/5 <==
============
2.75 / -2.2 =
- 1.25 <==
============
(-2 3/5) / (3/5) =
- 13/5 * 5/3 =
-13/3 or - 4 1/3 <==
============
(2 1/4) / (3/4) =
9/4 * 4/3 =
9/3 =
3 <==

7 0

3 years ago

Use substitution to solve the system of equations.

Katarina [22]

The colution to your problem is a. (3,-6)

5 0

4 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

3 years ago