Answer:
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.7291
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have 
. So
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
Answer:
Step-by-step explanation:
<u>Given</u>
<u>Cross multiply and solve for x:</u>
Answer:
t = 8.9
Step-by-step explanation:
6.35t = 56.515
56.515 / 6.35 = t
t = 8.9
Answer:
5x+4
Step-by-step explanation:
6x+8-x-4
5x+4
81.4% ≅ 81%. The probability that a customer ordered a hot drink given that he or she ordered a large is 81%.
The key to solve this problem is using the conditional probablity equation P(A|B) = P(A∩B)/P(B). Conditional probability is the probability of one event occurring with some relationship to one or more other events.
Similarly to the previous exercise, P(A∩B) is the probability that a customer order a large hot drink. So, P(A∩B) = 22/100 = 0.22
For P(B), is the probability that a customer order a large drink whether hot or cold. P(B) = 27/100 = 0.27
P(A|B) = 0.22/0.27 = 0.814
multiplying by 100%, we obtain 81.4%
