Answer:
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
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
.
That is z with a pvalue of
, so Z = 1.645.
Now, find the margin of error 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 34.09 - 3.25 = 30.84.
The upper end of the interval is the sample mean added to M. So it is 34.09 + 3.25 = 37.34.
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Can u take its pictures and post it bcuz it might be easer to answer
Part A:Bees: f ( t ) = 1,500 * 0.88^tFlowering plants: g ( t ) = 800 - 25 tPart B :f ( 6 ) = 1,500 * 0.88^6 = 696.6 ≈ 697
g ( 6 ) = 800 - 25 * 6 = 800 - 150 = 650Part C :f ( 7 ) = 1,500 * 0.88^7 = 613g ( 7 ) = 800 - 7 * 25 = 625f ( 7 ) ≈ g ( 7 )After approximately 7 months.