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.

it would be 6xy and -16xy.

let f (x ) is a function and c be a positive real number,   the graph of the function g (x ) = f (cx ) is represented as follows.

when, c > 1,   the function compresses it in the x - direction.

when, 0 < c < 1, the function stretches it.

the function is g ( x ) = (9x )2.

the heighest power of x in g ( x ) = (9x )2 is 2, so the parent function is square function.

the equation of parent function is f ( x ) = x 2 .

the function is g ( x ) = (9x )2 and the parent function is f ( x ) = x 2 .

so, the functions f and g have the following relationship.

g ( x ) = f ( 9x ).

compare the function g ( x ) = f ( 9x ) with g (x ) = f (cx ).

c = 9 ( > 1 ).

so, the function g (x ) = (9x )2 compresses it in the x - direction.

step-by-step explanation: