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lianna [129]
3 years ago
7

6 points and brainliest answer No.1

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0
The answer is B.) 14 :):):):)
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What is the median of the data set 10,7,9,9,4, 6?
lilavasa [31]

Answer:

8

Step-by-step explanation:

The median is the middle of the numbers in a given data set.

To find the median, line up all of the numbers from greatest to least.

4, 6, 7, 9, 9, 10

Now, find the middle of the numbers. Since there is an even number of numbers in the data set, find the two middle numbers.

4, 6, 7, 9, 9, 10

Next, find the average of these two numbers. (To do this, add the numbers and divide that sum by 2.)

7 + 9 = 16

16/2 = 8

Therefore, the median is 8.

4 0
3 years ago
The figures above are similar find the missing length show your work
Misha Larkins [42]

find the ratio ( scale factor ) of the 2 known sides

large box has length of 5, small box has length of 3

3 divided by 5 = 0.6

 so now multiply the known height of large box (3) by 0.6 to find x

3 * 0.6 = 1.8

x = 1.8 inches



please mark brainliest.

6 0
3 years ago
A. y ≤ -3x + 4<br>B. y ≥ -3x - 4<br>C. y ≥ -3x + 4<br>D. y ≤ -3x - 4 ​
pashok25 [27]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
each different letter of the word experiment is written in a card and placed in a hat. One card is drawn from the hat. How many
serg [7]
There would be 8 possible outcomes. e x p r i m n t.
6 0
3 years ago
Given the function f(x) = -5x²-x+ 20, find f(3).<br> O-28<br> O-13<br> O62<br> O64
olganol [36]

Answer: \Large\boxed{First~Choice.~-28}

Step-by-step explanation:

<u>Given function expression</u>

f(x) = -5x² - x + 20

Requirement of the question

Find the value of f(3)

<u>Substitute the values into the function</u>

f(x) = -5x² - x + 20

f(3) = -5(3)² - (3) + 20

<u>Simplify the exponents</u>

f(3) = -5(9) - 3 + 20

<u>Simplify by multiplication</u>

f(3) = -45 - 3 + 20

<u>Simplify by subtraction and addition</u>

f(3) = -48 + 20

\Large\boxed{f(3)=-28}

Hope this helps!! :)

Please let me know if you have any questions

5 0
1 year ago
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