You measure 49 turtles' weights, and find they have a mean weight of 80 ounces. Assume the population standard deviation is 6.1 ounces. Based on this, construct a 99% confidence interval for the true population mean turtle weight. Round your answers to 2 decimal places.

The 99% confidence interval for the true population mean turtle weight is between 77.76 and 82.24 ounces.

Step-by-step explanation:

We have to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a p-value of , so Z = 2.575.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 80 - 2.24 = 77.76 ounces.

The upper end of the interval is the sample mean added to M. So it is 80 + 2.24 = 82.24 ounces.

The 99% confidence interval for the true population mean turtle weight is between 77.76 and 82.24 ounces.

the answer will be the last one

answer: - (a)

(b) 88 seats

step-by-step explanation:

given: in an auditorium, there are 22 seats in the first row and 28 seats in the second row.

thus first term=

second term=

the number of seats in a row continues to increase by 6 with each additional row.

⇒d= 6

a) given formula :

b)   let the number of seats (n) = 12

substitute n=12 in the explicit rule formed in (a)

∴there are 88 seats in 12 rows.