Question 7(Multiple Choice Worth 1 points) (08.02 LC) Eva conducted a survey to find out the amount of time high schoolers spend
1 answer:
Answer:
The answer is "Option B".
Step-by-step explanation:
The difference between most time and also the least spending time on Internet surfing is 3 hours. Since we do not have charts for tables etc., only 3 can be used we need. A range is defined as the difference between the largest and the smallest amounts. The range between both the largest as well as the smallest is unique. In this reply, it tells us that the gap between most time and the fewer hours invested surfing the web is 3 hours.
- In option A, it is wrong since the range has nothing to do with formulas. (Of course, the dividend with a divisor results in a quotient). Only subtraction and not division may be achieved.
- In option C, when all surf for exactly one hour, it could take the largest time of 3 hours and 3 hours, the last time. Add it into the equation and the range of the data present would've been 0.
- In option D, It is erroneous even as the range is not the mean, and the mean seems to be the average. We search for both the range, not the mean.
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Applying our power rule gets us our first derivative,
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.