The sixth term is 22 and the common difference is six Ward the question is what is the 15th term
1 answer:
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THIS IS THE COMPLETE QUESTION BELOW;
The weekly salaries of sociologists in the United States are normally distributed and have a known population standard deviation of 425 dollars and an unknown population mean. A random sample of 22 sociologists is taken and gives a sample mean of 1520 dollars.
Find the margin of error for the confidence interval for the population mean with a 98% confidence level.
z0.10 z0.05 z0.025 z0.01 z0.005
1.282 1.645 1.960 2.326 2.576
You may use a calculator or the common z values above. Round the final answer to two decimal places.
Answer
Margin error =210.8
Given:
standard deviation of 425
sample mean x=1520 dollars.
random sample n= 22
From the question We need to to calculate the margin of error for the confidence interval for the population mean .
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
Answer:
a) P=0.0175
b) P=0.0189
Step-by-step explanation:
For both options we have to take into account that not only the chance of a "superevent" will disable both suppliers.
The other situation that will disable both is that both suppliers have their "unique-event" at the same time.
As they are, by definition, two independent events, we can calculate the probability of having both events at the same time as the product of both individual probabilities.
a) Then, the probability that both suppliers will be disrupted using option 1 is
b) The probability that both suppliers will be disrupted using option 2:
Pue = probability of a unique event
Pse = probability of a superevent
Answer:
a21 = -61
Step-by-step explanation:
(subtract to eliminate a₁)
9 = -3d
d = -3
-19 = a₁ + (6)(-3)
-1 = a
a21 = -1 + (21 - 1)(-3)
= -61