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AURORKA [14]
3 years ago
15

Help please I have no idea​

Mathematics
2 answers:
Vesna [10]3 years ago
8 0

Step-by-step explanation:

What the question is asking you to do is

convert

\frac{20}{200}

into a percentage.

First, we will divide the numerator and denominator by 2 to get the denominator to 100 , then from there, we can get the percentage value from the numerator.

\frac{20 \div 2}{200 \div 2}  \\  =  \frac{10}{100}

From the fraction above, we can see that the 20 is 10% of 200 , from the numerator.

Tamiku [17]3 years ago
4 0

Answer:

There is nothing here to answer.

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I need help finding the angles. The answer Isn't even
mart [117]

The 3 inside angles of a triangle equal 180.

Add the 3 angles together to equal 180:


X + (5x-20) + 4x = 180

Simplify the left side by adding like terms:

10x-20 = 180

Add 20 to each side:

10x = 200

Divide both sides by 10:

x = 20


Angle A = x = 20

Angle B = 4X = 4(20) = 80

Angle C = 5x-20 = 5(20) -20 = 100-20 = 80

6 0
3 years ago
What is 9/12 as a decimal
UNO [17]

Answer:

.75

Step-by-step explanation:

u just do 9÷12 to find a decimal

8 0
3 years ago
A variety of two types of snack packs are delivered to a store. The box plots compare the number of calories in each snack pack
adelina 88 [10]

OPTIONS:

A. The interquartile range of the trail mix data is greater than the range of the cracker data.

B. The value 70 is an outlier in the trail mix data.

C. The upper quartile of the trail mix data is equal to the maximum value of the cracker data.

D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

Answer:

D.

Step-by-step explanation:

With the given information about how the box plot looks like, let's examine each option to see if they are true or not.

<u><em>Option A: "The interquartile range of the trail mix data is greater than the range of the cracker data."</em></u>

The interquartile range of trail mix data = 105 - 90 = 15

Range of cracker data = 100 - 70 = 30

Option A is NOT TRUE.

<u><em>Option B: "The value 70 is an outlier in the trail mix data.</em></u>"

This is NOT TRUE. There are not outliers as 70 is the minimum value if the ranges of the data set for the trail mix.

<em><u>Option C: "The upper quartile of the trail mix data is equal to the maximum value of the cracker data."</u></em>

Upper quartile of the trail mix data = 105

Max value of cracker data = 100

This statement is NOT TRUE.

<em><u>Option D: "The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers."</u></em>

The greater the range value, the greater the variation. Thus,

Range value of the trail mix data = 115 - 70 = 45

Range value of the cracker data = 100 - 70 = 30

This is statement is correct because trail mix data have a greater range value, hence, it has a greater variation in the number of calories.

6 0
3 years ago
Read 2 more answers
Which number line represents the solution set for the inequality –4(x + 3) ≤ –2 – 2x?
Delicious77 [7]

Answer:

–4(x + 3) ≤ –2 – 2x

>>.....-4x -12 ≤ -2 -2x

>>  -12 +2 ≤ +2x  

>> - 10 ≤ 2x

>> -5 ≤ x............>> x >= -5  

Step-by-step explanation:

you didn't post any number lines...

3 0
3 years ago
The following frequency distribution shows the yearly tuitions (in $1,000s) of a sample of private colleges.
hoa [83]

Answer:

19.714

Step-by-step explanation:

Given :

Yearly Tuition Frequency

12-16 5

17-21 4

22-26 3

27-31 2

We calculate the midpoint (x) = (lower + upper) / 2

The mean yearly tuition :

The midpoint (x) :

X ___ F ___ Fx

14 __ 5

19 __ 4

24 __ 3

29 __ 2

The mean = Σfx / Σf

Σfx / Σf = (14*5 + 19*4 + 24*3 + 29*2) / (5+4+3+2)

Mean = 276 / 14

Mean yearly tuition = 19.714

8 0
3 years ago
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