Suppose that 2121% of all births in a certain region take place by Caesarian section each year. a. In a random sample of 500500 births, how many, on average, will take place by Caesarian section? b. What is the standard deviation of the number of Caesarian section births in a sample of 500500 births? c. Use your answers to parts a and b to form an interval that is likely to contain the number of Caesarian section births in a sample of 500500 births.

a)

b)

c) Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

So we expect about 86 and 123 most of the numbers of Caesarian section births

Step-by-step explanation:

For this case we can define the random variable X as the number of births in the  Caesarian section and from the data given we know that the distribution of X is:

Part a

The expected value for this distribution is given by:

Part b

The variance is given by:

And the deviation would be:

Part c

Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

So we expect about 86 and 123 most of the numbers of Caesarian section births

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points

the anwser is d) 75°

step-by-step explanation: