The given expression can be simplified in many ways by grouping like terms. The simplest form is obtained by factoring out a²b which gives us the following expression.
From the given diagram, you can see that the center of the hyperbola is placed at the origin, so first option is true (see attached diagram for definition of center, vertices, foci, i.e.)
There are two vertices of the hyperbola, they are placed at (-6,0) and (6,0), so second option is true.
The transverse axis is the segment connecting vertices, this segment is horizontal, so option 3 is false.
The foci are not placed within the rectangular reference box, so this option is false.
The directrices are vertical lines with equations , so this option is true.
And we can find this probability using excel or the normal standard tabe and we got:
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 temperatures of a population, and for this case we know the distribution for X is given by:
Where and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using excel or the normal standard tabe and we got:
The simplest fraction for is . Write the upper bound as a fraction with the same denominator:
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Hence the range for would be:
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If the denominator of is also , then the range for its numerator (call it ) would be . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than .
To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)
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At this point, the difference between the numerators is now . That allows a number ( in this case) to fit between the bounds. However, can't be written as finite decimals.
Try multiplying the numerator and the denominator by a different number.
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It is important to note that some expressions for can be simplified. For example, because of the common factor .