Answer:
x = 16
Step-by-step explanation:
x
-- = 9-1 ...... first shift 1 and it will be -1
2
x = 8 ....... secondly 9-1 =8
--
2
x = 16 ..... criss cross and 2×8 = 16
there fore the value of x is equals to 16.
you can check it!
Answer:
Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel
And the anfle is approximately 
Step-by-step explanation:
For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.
a=[1,2,-2], b=[4,0,-3,]
The dot product on this case is:
Since the dot product is not equal to zero then the two vectors are not orthogonal.
Now we can calculate the magnitude of each vector like this:
And finally we can calculate the angle between the vectors like this:
And the angle is given by:
If we replace we got:
Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel
And the anfle is approximately
Answer: I'm guessing it would be a parabola, with the line going through -6 on the y-axis and passing through 2 and 6 on the x-axis, but we cannot see any answers, so therefore we can't answer it accurately.
Step-by-step explanation:
Divide the current with the previous
18/15 = 1.2
move the decimal point to the right two places to get the percentage
120% is your answer
hope this helps
Answer:
a) 3.128
b) Yes, it is an outerlier
Step-by-step explanation:
The standardized z-score for a particular sample can be determined via the following expression:
z_i = {x_i -\bar x}/{s}
Where;
\bar x = sample means
s = sample standard deviation
Given data:
the mean shipment thickness (\bar x) = 0.2731 mm
With the standardized deviation (s) = 0.000959 mm
The standardized z-score for a certain shipment with a diameter x_i= 0.2761 mm can be determined via the following previous expression
z_i = {x_i -\bar x}/{s}
z_i = {0.2761-0.2731}/{ 0.000959}
z_i = 3.128
b)
From the standardized z-score
If [z_i < 2]; it typically implies that the data is unusual
If [z_i > 2]; it means that the data value is an outerlier
However, since our z_i > 3 (I.e it is 3.128), we conclude that it is an outerlier.
