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n200080 [17]
3 years ago
8

Find the median and mean of the data set below:

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer:

Step-by-step explanation:

Mean=sum of data/no of data

=161/7

=23

median

first of all arrange the data frome ascending to descending order

4,14,18,23,28,35,39

median=(N+1)/2(N=no of data)

=7+1/2

=8/2

=4 th term

therefore median = 23

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The first term of an arithmetic sequence is 5. if the25th term is 101 find the common difference?
d1i1m1o1n [39]
An arithmetic sequence is one where each term is a constant difference, called the common difference, from the preceding term.  The arithmetic sequence can always be expressed as:

a(n)=a+d(n-1), a=first term, d=common difference, n=term number.

We are given two terms and term numbers, so we can solve for the common difference...

101=5+d(25-1)

101=5+25d-d

101=5+24d

96=24d

d=4

So the common difference is 4.


8 0
3 years ago
Read 2 more answers
A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

6 0
3 years ago
Marty divided an 8-pound bag of potatoes into a 7 equal piles.Which rational number represents the weight of the potatoes in eac
erma4kov [3.2K]
I think the answer is 1.14 but I'm not sure.
4 0
3 years ago
Given f(x) = 3x^3 + kx – 5, and x + 1 is a factor of f(x), then what is
DENIUS [597]

Answer:

Step-by-step explanation:

k=-6 ok

5 0
3 years ago
What is the measure of x?
Pavlova-9 [17]
We know angle BAD= CBA so we can do:
42+105 + x = 180
x = 180-105-42
x=33

Hope this helps!! :)
8 0
3 years ago
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