It might be Annual salary?
Answer:
The 95% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).
Answer: The answer is 88
Step-by-step explanation: It is recursive because you can add 78+10=88
Circumference of entire circle = 2 * PI * radius
2 * PI * 3 =
<span>
<span>
<span>
18.8</span></span></span>5 meters
a 30 degree arc is (30/360) or 1/12 of a circle.
So the arc equals (18.85 meters / 12) or
<span>
<span>
<span>
1.57</span></span></span> meters.
Answer:

Step-by-step explanation: