A survey found that 30% of teenagers don’t wear a seatbelt. If five teenagers are selected at random, find the probability that
1 answer:
You might be interested in
Answer:
The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 67 - 4.54 = 62.46 ounches.
The upper end of the interval is the sample mean added to M. So it is 67 + 4.54 = 71.54 ounces.
The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.
Answer:
A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.
Step-by-step explanation:
A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):
The process will help us to understand what is happening, at first we rewrite the function:
And then, we use the chain rule to differentiate:
Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).
In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):
if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b) (and below we explain why this happen)
Considering that ln(a) is a growing function and ln(e)=1, we have:
if a<e<b, then ln(a)< 1 <ln(b)
if a<e, then aˣln(a)<aˣ
if e<b, then bˣ<bˣln(b)
And because eˣ is defined to be the same as its derivative, the cases above results in the following
if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)
if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )
if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )
but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)
