Answer: 
Step-by-step explanation:
As per given , we have
n = 11

Since population standard deviation is missing, so we use t-test.
Critical t-value for 99% confidence :
[using two-tailed t-value table]
Confidence interval :

Hence, 99% confidence interval for the mean amount of time that students spend in the shower each day.= 
Answer:
a = 21
b = 63
c = 42√3
d = 21√3
Step-by-step explanation:
The sides of a 30°-60°-90° triangle have the ratios 1 : √3 : 2. The given side (42) is the longest side of the smallest triangle, and the shortest side of the largest triangle.
That means the other sides of the smallest triangle will be ...
a = 42/2 = 21
a+b = 2(42) = 84
b = (a+b) -a = 84 -21 = 63
d = 21√3 . . . . middle-length side of the smallest triangle
c = 42√3 . . . . middle-length side of the largest triangle
The values of the variables are ...
- a = 21
- b = 63
- c = 42√3
- d = 21√3
Answer:
The two angles mentioned are alternate interior angles, so they are congruent.
303 feet.
Step-by-step explanation:
If you draw off the flag you can see they are alternate interior and are congruent.
Also again, make a diagram it really helps a lot
:)
Answer:
11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they received a pneumococcal vaccination, or they did not. The probability of an adult receiving a pneumococcal vaccination is independent of other adults. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
70% of U.S. adults aged 65 and over have ever received a pneumococcal vaccination.
This means that 
20 adults
This means that 
Determine the probability that exactly 12 members of the sample received a pneumococcal vaccination.
This is P(X = 12).


11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.