Astudy of 35 golfers showed that their average score on a particular course was 92. the sample standard deviation was 5. we want to construct a 95% confidence interval to estimate the value of the true population mean. what is the t-value that should be used to compute the confidence interval (to the third decimal place)?

Step-by-step explanation:

We have a sample of size n = 35, and . As we want to construct a 95% confidence interval to estimate the value of the true population mean , we should use the pivotal quantity given by which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value , i.e., regarding the t-distribution with 34 degrees of freedom, we have an area of 0.025 above 2.0322 and below the probability density function. The rationale behind this is that because of the simmetry of the t distribution.

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: