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Mariulka [41]
3 years ago
5

What is the main advantage of using money instead of bartering? A. Many more people today use money than bartering to get goods

and ervices. B. Not all countries use the same types of money in their economies C. Money comes in different values to ay for goods and services D. people do not have to have a good or a service that another person wants
Mathematics
2 answers:
sladkih [1.3K]3 years ago
7 0
Yes, I agree! The answer is D!
alekssr [168]3 years ago
6 0

D. people do not have to have a good or service that another person wants

Reason: Money gives people a universal unit of value. For example, If I needed a toothbrush and I have a pair of scissors, In order for me to get a tooth brush is to trade my scissors with someone else who has a tooth brush and wants scissors. As you can see, this isn't practical. If I had money, I will only need to find someone who has a tooth brush.

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Philip saves $8 each month. How many months will it take him to save at least $60?
mars1129 [50]
7.5 months for him to save up to 60 ;)
8 0
4 years ago
Read 2 more answers
Find the midpoint of the segment with the following endpoints.<br> (2,9) and (8,1)
vazorg [7]

Answer:

\displaystyle (5,5)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Coordinates (x, y)
  • Midpoint Formula: \displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Step-by-step explanation:

<u>Step 1: Define</u>

Point (2, 9)

Point (8, 1)

<u>Step 2: Identify</u>

(2, 9) → x₁ = 2, y₁ = 9

(8, 1) → x₂ = 8, y₂ = 1

<u>Step 3: Find Midpoint</u>

Simply plug in your coordinates into the midpoint formula to find midpoint

  1. Substitute in points [Midpoint Formula]:                                                         \displaystyle (\frac{2+8}{2},\frac{9+1}{2})
  2. [Fractions] Add:                                                                                                  \displaystyle (\frac{10}{2},\frac{10}{2})
  3. [Fractions] Divide:                                                                                              \displaystyle (5,5)
7 0
3 years ago
This problem is lowkey tricky. Can someone help me out
notka56 [123]

Answer:

x = 10

Step-by-step explanation:

Proof: If a line is parallel to one side of the triangle, then it divides the other two sides proportionally.

6/9 = x/15

9x = 90

x = 10

6 0
2 years ago
Q Searchmatics Gr 7)I need 4 cups of flours and 2 cups of sugar to make a cake. How many cups of sugar do I need if Ihave 12 cup
Ede4ka [16]

You will need 6 cups of sugar.

Step - by - Step Explanation

What to find? <em>The number of cups of sugar needed to make a cake with 12 cups of flour.</em>

<em />

Given:

Cups of flours = 4

Cups of sugar = 2

Cups of flour =12

Let x be the number of cups of sugar needed to make a cake with 12 cups of flour.

Using proportion;

4 cups of flour = 2 cups of sugar

12 cups of flour = x

Cross - multiply

4x = 24

Divide both-side of the equation by 4

\frac{\cancel{4}x}{\cancel{4}}=\frac{24}{4}

x =6

Hence, you will need 6 cups of sugar.

6 0
1 year ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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