One number is two greater than another—the product of the numbers 1s 143. Find the numbers. One pair of numbers, both of which a
1 answer:
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Answer:
And if we solve for a we got
And the limits for this case are: (214.4; 295.4)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:
Where and
The confidence level is 95.44 and the signficance is and the value of
. And the critical value for this case is
Using this condition we can find the limits
And if we solve for a we got
And the limits for this case are: (214.4; 295.4)
The graph from which the position of the point <em>A</em> can determined following
the multiplication with a scalar is attached.
Responses:
- If <em>A</em> is in quadrant I and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant III</u>
- If A is in quadrant II and is multiplied by a positive scalar, <em>c</em>, then c·A is in <u>quadrant II</u>
- If <em>A</em> is in quadrant II and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant IV</u>
- If <em>A</em> is in quadrant III and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant I</u>
Background information;
The question relates to the coordinate system with the abscissa represent the real number and the ordinate representing the imaginary number.
Solution:
If A is in quadrant I; A = a + b·i
When multiplied by a negative scalar, <em>c</em>, we get;
c·A = c·a + c·b·i
Therefore;
c·a is negative
c·b is negative
- c·A = c·a + c·b·i is in the <u>quadrant III</u> (third quadrant)
If A is quadrant II, we have;
A = -a + b·i
When multiplied by a positive scalar <em>c</em>, we have;
c·A = c·(-a) + c·b·i = -c·a + c·b·i
-c·a is negative
c·b·i is positive
Therefore;
- c·A = -c·a + c·b·i is in <u>quadrant II</u>
Multiplying <em>A</em> by negative scalar if <em>A</em> is in quadrant II, we have;
c·A = -c·a + c·b·i
-c·a is positive
c·b·i is negative
Therefore;
c·A = -c·a + c·b·i is in <u>quadrant IV</u>
If A is in quadrant III, we have;
A = a + b·i
a is negative
b is negative
Multiplying <em>A</em> with a negative scalar <em>c</em> gives;
c·A = c·a + c·b·i
c·a is positive
c·b is positive
Therefore;
- c·A = c·a + c·b·i is in<u> quadrant I</u>
Learn more about real and imaginary numbers here;
