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Mandarinka [93]
3 years ago
14

If you wanted the item that cost the most at the top of the table, you would sort in descending order. true false

Mathematics
2 answers:
Leya [2.2K]3 years ago
6 0
It is true because u need to sort it so u can find the one that cost the most.

Yanka [14]3 years ago
6 0

Answer:

TRUE

Step-by-step explanation:

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A stock investment went up $25\%$ in 2006. Starting at this increased value, what percent would it have to go down in 2007 to be
Sladkaya [172]

Answer:

20%

Step-by-step explanation:

Let x represent the initial value of the stock at the beginning of 2006. And y the value at the end of 2006.

When it went up by 25% in 2006;

y = x + 25% of x = x + 0.25x = 1.25x

For the value to go back to the original price, it has to decrease from y to x;

y = 1.25x

The percentage decrease from y to x is;

= (y-x)/y × 100%

Substituting the values;

= (1.25x -x)/1.25x × 100%

= 0.25x/1.25x × 100%

= 0.2 × 100%

= 20%

Therefore, it will have to go down 20%

6 0
4 years ago
1 y = 3x - 4<br> 2 x + y = 8<br> use the method of substitution
Charra [1.4K]

Answer:

x = 2.4; y = 3.2

Step-by-step explanation:

y = 3x - 4

2x + y = 8

2x + 3x - 4 = 8

5x - 4 = 8

5x = 12

x = 12 / 5

x = 2.4

y = 3(2.4) - 4

y = 7.2 - 4

y = 3.2

5 0
2 years ago
PLEASE HELP!
Luba_88 [7]
2x - 2y = 2

2x - 2 x 0 = 2

2x - 0 = 2

2x = 2

x = 1

2x - 2y = 2

2 x 0 - 2y = 2

y = - 1
3 0
3 years ago
Read 2 more answers
Musical megabytes How much disk space
alexdok [17]

Given:

Consider the file sizes (in  megabytes) for 18 randomly selected files on  Gabriel's mp3 player are shown in the below figure.

To find:

The outliers in the  distribution.

Solution:

We have, the given data set

2.4, 2.7, 1.6 , 1.3 , 6.2, 1.3, 5.6, 1.1, 2.2, 1.9, 2.1, 4.4, 4.7, 3.0, 1.9, 2.5, 7.5, 5.0

Arrange the data in ascending order.

1.1, 1.3, 1.3, 1.6, 1.9, 1.9, 2.1, 2.2, 2.4, 2.5, 2.7, 3.0, 4.4, 4.7, 5.0, 5.6, 6.2, 7.5,

Divide the data in 4 equal parts.

(1.1, 1.3, 1.3, 1.6), 1.9, (1.9, 2.1, 2.2, 2.4),( 2.5, 2.7, 3.0, 4.4), 4.7, (5.0,5.6, 6.2, 7.5)

It is clear that,

Q_1=1.9

Q_3=4.7

IQR=Q_3-Q_1

IQR=4.7-1.9

IQR=2.8

Now,

Interval=[Q_1-1.5IQR,Q_3+1.5IQR]

Interval=[1.9-1.5(2.8),4.7+1.5(2.8)]

Interval=[-2.3,8.9]

All the data values lies in the interval [-2.3,8.9]. Therefore, the given data set have no outliers.

7 0
3 years ago
Kayla has 1\frac{3}{7} cups of yogurt to make smoothies. Each smoothie uses \frac{1}{4} cup of yogurt. What is the maximum numbe
Fiesta28 [93]

Answer: The maximum number of smoothies is 6 cups

Step-by-step explanation: The quantity of yogurt Kayla has is 1³/₇ cups and she needs to make smoothies that would require, 1/4 cup. The number of smoothies she can make from this available quantity can be mathematically expressed as follows;

Number of smoothies = Total Quantity/Quantity per cup

So if we represent Total quantity (1 ³/₇ cups) by x and the quantity per cup (1/4) by y, then the number of smoothies she can make from her available amount of yogurt can be expressed as follows;

Number of smoothies = x/y

The reason is simple, Kayla can determine how many cups of smoothies she can make provided she knows how much quantity each cup of smoothie would require. That means, if for example she has 10 cups of yogurt and each smoothie requires 5 cups, she simply needs to find out how many 5 cups she can get out of 10 cups, which now translates into 10 divided by 5. Similarly, she has 1 ³/₇ cups of yogurt and to determine how many ¹/₄ cups can come out of this she simply needs to divide the total quantity of yogurt by the amount each cup requires, which bring us back to the equation;

Number of smoothies = x/y

Where x = 1 ³/₇ cups of yogurt and y = ¹/₄ cup of yogurt

Number of smoothies = 1 ³/₇÷ ¹/₄

Number of smoothies = (¹⁰/₇) x (⁴/₁)

Number of smoothies = 40/7

Number of smoothies = 5 ⁵/₇

The maximum number of smoothies Kayla can make is 6 cups, because the result shows 5 cups and a fraction which is greater than half a cup. Therefore the last one that measures 5/7 of a cup shall be the sixth one.

8 0
3 years ago
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