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3 years ago

What is the median of this data set 14,18,31,34,44,50

Mathematics

1 answer:

DochEvi [55]3 years ago

6 0

The median is 31.83

Because:

14 + 18 + 31 + 34 + 44 + 50 = 191

191 / 6 = 31.83 (the median)

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Here are the shopping times (in minutes) of ten shoppers at a local grocery store.

d1i1m1o1n [39]

The frequency which is to filled in the table is 3,1,2,4.

Given shopping time of ten shopkeeper: 36,20,32,18,20,33,28,34,37,23.

We have to complete the frequency in the table. Frequency of something is the number of times that number is coming.

Shopping time is given as 36,20,32,18,20,33,28,34,37,23.

Intervals for which frequency is given is 18 to 22, 23 to 27, 28 to 32, 33 to 37.

Numbers in the range 18 to 22=18,20,23 therefore 3

Numbers in the range 23 to 27=23 therefore 1

Numbers in the range 28 to 32=32 , 28 therefore 2

Numbers in the range 33 to 37=36, 33, 37, 34 therefore 4.

Hence the frequency which is to filled are 3,1,2,4.

Learn more about frequency at brainly.com/question/254161

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8 0

2 years ago

Find the missing side length in the image below

tatuchka [14]

Find the area and volume of the trapezoids first

3 0

3 years ago

Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example

polet [3.4K]

Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

$\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{cc}e&f\\g&h\end{array}\right] = \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]$

For the product to be a diagonal matrix,

a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g

Consider the following sets of values

$a=1, \ \ b=2, \ \ c=3, \ \ d = 4, \ \ e=\frac{1}{3}, \ \ f=-1, \ \ g=-\frac{1}{4}, \ \ h=\frac{1}{2}$

The the matrix product becomes:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]$

Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.

8 0

3 years ago

Read 2 more answers

2. Calculate an expression for dy/dx and d2y/dx2 in terms of t if the parametric pair is given as tan(x) = e^at and e^y = 1 + e^

Ber [7]

I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives

sec²(x) dx/dt = a exp(at)

Recall that

sec²(x) = 1 + tan²(x)

Then we have

(1 + tan²(x)) dx/dt = a exp(at)

(1 + exp(2at)) dx/dt = a exp(at)

dx/dt = a exp(at) / (1 + exp(2at))

If exp(y) = 1 + exp(2at), then differentiating with respect to t yields

exp(y) dy/dt = 2a exp(2at)

(1 + exp(2at)) dy/dt = 2a exp(2at)

dy/dt = 2a exp(2at) / (1 + exp(2at))

By the chain rule,

dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)

Then the first derivative is

dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))

dy/dx = (2a exp(2at)) / (a exp(at))

dy/dx = 2 exp(at)

Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write

dy/dx = f(t)

By the chain rule, the derivative is

d²y/dx² = df/dx

d²y/dx² = df/dt • dt/dx

d²y/dx² = (df/dt) / (dx/dt)

d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))

d²y/dx² = 2 (1 + exp(2at))

4 0

2 years ago

what is the equation in point slope form of the line that passes through the points (2,-3) and (-5,8)

USPshnik [31]

Answer:

y+3 = -11/7x(x-2)

Step-by-step explanation:

4 0

3 years ago