In an effort to estimate the mean amount spent per customer for dinner at major Atlanta restaurant, data were collected for a sample of 49 customers over a period of three weeks. Assume a population standard deviation of $2.50. If the sample mean is $22.60, compute a 98% confidence interval for population mean?

Confidence Interval is (21.76928571,23.43071429)

Step-by-step explanation:

The Formula we are going to use is:

±

Where:

μ is the mean of sample

σ is the standard deviation of population

n is the sample size

z is the distribution

In our case:

μ=$22.60,σ=$2.50

For 98% Confidence level, 1-α=98%

α=2%

α/2=1%=0.01

From Cumulative standardized Normal Distribution table

=2.326

P=23.430714      P=21.769286

Confidence Interval is (21.76928571,23.43071429)

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

where r the answers

step-by-step explanation: