Please help? Is this correct i need tight now poease
1 answer:
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For this case, what we must do is calculate the mean of the given data.
For this, we must do the following:
Add the given numbers and divide by the number of numbers.
We have then:
From here, we clear the value of x.
We have then:
Answer:
Tasha would have to score 90 on the sixth test to have a mean of 85
For this, we must do the following:
Add the given numbers and divide by the number of numbers.
We have then:
From here, we clear the value of x.
We have then:
Answer:
Tasha would have to score 90 on the sixth test to have a mean of 85
This can be solved a couple of ways. One way is to use the Pythagorean theorem to write equations for the magnitude from the components of the forces. That is what was done in the graph here.
Another way is to use the Law of Cosines, which lets you make direct use of the angle between the vectors.
.. 13 = a^2 +b^2 -2ab*cos(90°)
.. 19 = a^2 +b^2 -2ab*cos(120°)
Subtracting the first equation from the second, we have
.. 6 = -2ab*cos(120°)
.. ab = 6
Substituting this into the first equation, we have
.. 13 = a^2 +(6/a)^2
.. a^4 -13a^2 +36 = 0
.. (a^2 -9)(a^2 -4) = 0
.. a = ±3 or ±2
The magnitudes of the two forces are 2N and 3N, in no particular order.
Another way is to use the Law of Cosines, which lets you make direct use of the angle between the vectors.
.. 13 = a^2 +b^2 -2ab*cos(90°)
.. 19 = a^2 +b^2 -2ab*cos(120°)
Subtracting the first equation from the second, we have
.. 6 = -2ab*cos(120°)
.. ab = 6
Substituting this into the first equation, we have
.. 13 = a^2 +(6/a)^2
.. a^4 -13a^2 +36 = 0
.. (a^2 -9)(a^2 -4) = 0
.. a = ±3 or ±2
The magnitudes of the two forces are 2N and 3N, in no particular order.