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zzz [600]
3 years ago
15

Need help with the answer

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
3 0
The answer is 50.24 inches :))
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What are the mean, median, mode, and range of the data set given the altitude of lakes in feet: -11, -28, -17, -25, -28, -39, -6
GrogVix [38]

Answer:

Mean=-21.5

Median=-26.5

Mode=-28

Range= -6 to-46

Step-by-step explanation:

Arranging the numbers given either in ascending or descending order. In this case, I used descending order as follows

-6

-11

-17

-25

-28

-28

-39

-46

Mode is the number that occurs more than others. In this case, only -28 appears twice hence it is the mode

The mean is given by adding all the numbers and dividing by the frequency.

The sum of numbers will be -6+-11+-17+-25+-28+-28+-39+-46=-172

Mean=-172/8=-21.5

The median is the number that appears in the middle after arranging the numbers in descending order as above. In this case, two numbers appear in the middle, that is -25 and -28 hence median -25+-28/2=-26.5

The range of numbers id from -6 to-46 as arranged above

8 0
3 years ago
Find the length of the third side. If necessary, round to the nearest tenth.<br> 17<br><br> 25
Flura [38]

Answer:

the real answer is 48 but if you round it to the nearest tenth it's 50

Step-by-step explanation: I hope my answer helps, if it does help can I get the brainliest answer, thanks.

5 0
3 years ago
Read 2 more answers
PLLLLLSSSS Heeeellppp ill mark brainliest I promisseeeeeeee plsssss
WARRIOR [948]

Answer:

2\,(x-1)^2=4

which is the first option in the list of possible answers.

Step-by-step explanation:

Recall that the minimum of a parabola generated by a quadratic expression is at the vertex of the parabola, and the formula for the vertex of a quadratic of the general form:

y=ax^2+bx+c

is at   x_{vertex}=\frac{-b}{2\,a}

For our case, where a=2\,\,, b=-4\,\,,\,\,and \,\,c=-2  we have:

x_{vertex}=\frac{-b}{2\,a}\\x_{vertex}=\frac{4}{2\,*\,2}\\x_{vertex}=1

And when x = 1, the value of "y" is:

y(x)=2x^2-4x-2\\y(1)=2(1)^2-4(1)-2\\y(1)=2-6\\y(1)=-4\\y_{vertex}=-4

Recall now that we can write the quadratic in what is called: "vertex form" using the coordinates (x_{vertex},y_{vertex)of the vertex as follows:

y-y_{vertex}=a\,(x-x_{vertex})^2

Then, for our case:

y-(-4)=2\,(x-1)^2\\y=2\,(x-1)^2-4

Then, for the quadratic equal to zero as requested in the problem, we have:

y=2\,(x-1)^2-4=0\\2\,(x-1)^2-4=0\\2\,(x-1)^2=4

5 0
3 years ago
Solve for X.<br><br> Y=2/3×-4
ArbitrLikvidat [17]

Answer: -8/3

Step-by-step explanation:

Multiplying a positive and a negative equals negative: (+)•(-)=(-)

y= - 2/3•4

Y= -8/3

3 0
4 years ago
Please help me with this I don’t understand
Radda [10]

Decreased by means " -".

11.4 -15.7 = -4.3

Answer A.

7 0
4 years ago
Read 2 more answers
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