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2 years ago

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McDonald's is the king of the burger world, as it has more hamburger joints than any other restaurant chain. In 2013, McDonald's

had 1427 locations in Canada. If the population of Canada was approximately 35.16 million citizens in 2013, what was the number of McDonald's restaurants per capita in Canada in 2013? Express your answer in scientific notation rounded correctly to three decimal places.

1 answer:

Finger [1]2 years ago

4 0

Sent you a picture of the solution.

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Find the difference of -3/5 - (-5/9)<br> 20 points y'all<br><br> thanks<br> -Throckmorton

AlladinOne [14]

The work to your question

6 0

3 years ago

What is x and y math help

Readme [11.4K]

<h3>Answer:</h3>

20 cans of cola
10 cans of root beer

<h3>Step-by-step explanation:</h3>

x and y are whatever you want them to be.

It can be convenient for solving a problem like this to use x and y to represent <em>what the problem is asking for</em>: the number of cans of cola and the number of cans of root beer. It is also convenient (less confusing) to use those variable names in the same order that the nouns of the problem are named:

... x = # of cans of cola

... y = # of cans of root beer

Then the problem statement tells you ...

... x + y = 30 . . . . . . . 30 cans total were bought

... x = 2y . . . . . . . . . . the number of cans of cola is twice the number of cans of root beer

_____

This set of equations is nicely solved by substitution: use the second equation to substitute for x in the first.

... (2y) +y = 30 . . . . . put 2y where x was

... 3y = 30 . . . . . . . . collect terms

... y = 10 . . . . . . . . . divide by 3

... 2y = x = 20

<em>You're not done yet. You need to answer the question the problem asks.</em>

Jared bought 20 cans of cola and 10 cans of root beer.

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<em>Comment on x and y</em>

You customarily see x and y as the variables of a problem. Personally, I like to use variables that remind me what they stand for. In this problem, I might use "c" for cans of cola and "r" for cans of root beer. Then when I've found the solution, I know exactly how it relates to what the question is asking.

Always start by writing down what the variables stand for (as we did here). Sometimes, this is called <em>writing a Let statement</em>: <u>Let</u> x = number of colas; <u>let</u> y = number of root beers.

<em>Comment on problems of this type</em>

When a proportional relationship is given between the items in a sum (2 cola cans for every root beer can), it is often convenient to work the problem in terms of groups of items. Here, a group of 3 items can consist of 2 cola cans and 1 root beer can. Then 30 items will be 10 groups, so 10 root beers and 20 colas. The problem is solved even before you can name the variables.

Even when the relationship isn't exactly proportional, you can add or subtract the extras and still work the problem this way. Had we said colas numbered 3 more than twice as many root beers, we could have our groups of 3 total 27 (30 less the 3 extra), giving 9 root beers and 21 colas (3 + 2·9).

8 0

3 years ago

A party mix requires 4 cups of cereal for every 8 bowls of mix. How much party mix is required per bowl

jeka57 [31]

Answer:

0.25 party mix.

Step-by-step explanation:

4cups=8

2=4

1=2

1/2/2=0.25

4 0

1 year ago

Juans three math quizzes this week took him 1/3?4/6?and1/5 hour to complete. List all the fraction from least to greatest

Alex

Answer:

$\frac{1}{5}$ , $\frac{1}{3}$ , $\frac{4}{6}$ .

Step-by-step explanation:

Given fractions $\frac{1}{3}$ , $\frac{4}{6}$ , $\frac{1}{5}$ .

We need to list them from least to greatest.

In order to arrange them from least to greatest, we need to find the least common denominator of $\frac{1}{3}$ , $\frac{4}{6}$ , $\frac{1}{5}$ .

We have 3, 6 and 5 in denominators.

Least common multiple of 3, 6 and 5 is = 30, because 30 is least number that can be divided by all three number 3, 6 and 5.

Let us covert each denominator as 30.

Multiplying first fraction $\frac{1}{3}$ by 10 in top and bottom, we get

$\frac{1}{3} = \frac{1\times10}{3\times10}=\frac{10}{30}$

Multiplying first fraction $\frac{4}{6}$ by 5 in top and bottom, we get

$\frac{4}{6} = \frac{4\times5}{6\times5}=\frac{20}{30}$

Multiplying first fraction $\frac{1}{5}$ by 6 in top and bottom, we get

$\frac{1}{5} = \frac{1\times6}{5\times6}=\frac{6}{30}.$

Now, we can check $\frac{10}{30}, \frac{20}{30} \ and \ \frac{6}{30}.$

$\frac{6}{30}$ is the smallest, $\frac{10}{30}$ is greater and $\frac{20}{30}$ is the greatest.

Therefore, we can arrange fractions $\frac{6}{30}, \frac{10}{30} \ and \ \frac{20}{30}.$

Writing original fractions in place of equivalent fractions, we can write

$\frac{1}{5}$ , $\frac{1}{3}$ and $\frac{4}{6}$ .

Therefore, the order the amounts of paint from least to greatest is:

$\frac{1}{5}$ , $\frac{1}{3}$ , $\frac{4}{6}$ .

4 0

2 years ago

Changle. Show all work. Round each length to the nearest tenth and each angle to the

Sedaia [141]

Answer:

Part 1) $BC=12.2\ units$

Part 2) $m\angle A=55^o$

Part 3) $m\angle C=35^o$

Step-by-step explanation:

Part 1) Find AC

we know that

In the right triangle ABC of the figure

Applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$AC^2=7^2+10^2$

$AC^2=149\\AC=12.2\ units$

Part 2) Find the measure of angle A

we know that

In the right triangle ABC

$tan(A)=\frac{BC}{AB}$ ----> by TOA (opposite side divided by the adjacent side)

substitute the values

$tan(A)=\frac{10}{7}$

using a calculator

$m\angle A=tan^{-1}(\frac{10}{7})=55^o$

Part 3) Find the measure of angle C

we know that

In the right triangle ABC

$m\angle A+m\angle C=90^o$ ----> by complementary angles

substitute the given value

$55^o+m\angle C=90^o$

$m\angle C=90^o-55^o=35^o$

8 0

3 years ago