Answer:
The correct option is;
d. 24.2 to 25.6
Step-by-step explanation:
Here we have a sample with unknown population standard deviation, we therefore apply the student t distribution at 64 - 1 degrees of freedom
Therefore, we have
Where:
= Mean = 25
σ = Standard deviation = 2
n = Sample size = 64
t = T value at 98% =
Which gives
That is the value is from
24.40325 to 25.59675 which gives,by rounding to one decimal place, is
24.4 to 25.6.
Answer:
A. right 2, up 3
Step-by-step explanation:
We have that,
The function is transformed to
.
We see that,
The function f(x) is translated 2 units to the right and 3 units upwards to obtain the function g(x).
So, the correct transformation is 'right 2, up 3'.
Hence, option A is correct.
This is something you'll need a T table for, or a calculator that can compute critical T values. Either way, we have n = 10 as our sample size, so df = n-1 = 10-1 = 9 is the degrees of freedom.
If you use a table, look at the row that starts with df = 9. Then look at the column that is labeled "95% confidence"
I show an example below of what I mean.
In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26
<h3>Answer: 2.26</h3>
Answer:
A. Initially, there were 12 deer.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. After 15 years, there will be 410 deer.
D. The deer population incresed by 30 specimens.
Step-by-step explanation:
The amount of deer that were initally in the reserve corresponds to the value of N when t=0
A. Initially, there were 12 deer.
B.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C.
C. After 15 years, there will be 410 deer.
D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:
ΔN=N(15)-N(10)
ΔN=410 deer - 380 deer
ΔN= 30 deer.
D. The deer population incresed by 30 specimens.