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3 years ago

Find the median of this set of data 34 36 30 37 40 30 42 a 37 b 35 c 36 d 30

Mathematics

1 answer:

shusha [124]3 years ago

3 0

Answer:

Step-by-step explanation:

The median is the value that separates the higher half from the bottom half of the data sample (middle value).

We have

34, 36, 30, 37, 40, 30, 42

Write from the smallest to the largest number

30, 30, 34, 36, 37, 40, 42

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Choose a method . Then find the product 10x15

34kurt

Answer:

150

Step-by-step explanation:

15*10

Lets get rid of the 0 for now

15*1

That is 15 added to itself 0 times so it is 15

Now lets put the 0 back

and we will get

150

5 0

2 years ago

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WILL REWARD BRAINLIEST (40 POINTS PLEASE HELP)

Luba_88 [7]

Answer:

first of all I can't see nothing

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2 years ago

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Given the parent function of f(x) = x3, what change will occur when the function is changed to f(x) − 3?

DedPeter [7]

For this case we have that the main function is given by:

$f (x) = x ^ 3$

We apply the following transformation:

Vertical displacements:

Assume k> 0

To graph y = f (x) - k, move the graph k units down.

For k = 3 we have:

$y = f (x) - 3\\y = x ^ 3 - 3$

Answer:

when the function is changed to f (x) - 3 the change that will occur is:

Vertical displacement 3 units down.

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2 years ago

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Can someone tell me the points on a graph when you graph y=-5x

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Answer:

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Step-by-step explanation:

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2 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

3 years ago