You want to create an 80% confidence interval for the average age of people who attend U of O football games. You take a sample of 100 attendees and find the average age to be 43.7 years old with a standard deviation of 7 years. Find the value of t* for this confidence interval. Do not round your answer. Write your answer in decimal form, not as a fraction or percent.

80% confidence interval for the average age of people who attend U of O football games is [42.795 , 44.604].

Step-by-step explanation:

We are given that a sample of 100 attendees and find the average age to be 43.7 years old with a standard deviation of 7 years.

So, the pivotal quantity for 80% confidence interval for the population average start up cost is given by;

          P.Q. = ~

where, = sample average age = 43.7 years old

             = sample standard deviation = 7 years

            n = sample of attendees = 100

             = population average age of people

So, 80% confidence interval for the average age of people, is ;

P(-1.2915 < < 1.2915) = 0.80

P(-1.2915 < < 1.2915) = 0.80

P( < < ) = 0.80

P( < < ) = 0.80

80% confidence interval for = [ ,  ]

                                                 = [ , ]

                                                 = [42.795 , 44.604]

Therefore, 80% confidence interval for the population average age of people who attend U of O football games is [42.795 , 44.604].

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation:

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt