Let $\mu$ be the true (population) mean of statistics exam scores. We have a large random sample of n = 36 scores with a sample mean (sample mean score) of $\bar{x} = 68$ . We know that the population standard deviation is $\sigma = 3$ . A pivotal quantity is $(\bar{x}-\mu)/(3/\sqrt{36}) = (\bar{x}-\mu)/(3/6) = (68-\mu)/(1/2)$ which is approximately normally distributed and $P(-z_{0.05}\leq(68-\mu)/(1/2)\leq z_{0.05}) = 0.90$ where $-z_{0.05} =-1.6449$ , and so, $P(-1.6449\leq(68-\mu)/(1/2)\leq 1.6449) = 0.90$ . Therefore the 90% confidence interval is (68-(1/2)(1.6449), 68+(1/2)(1.6449)), i.e., ( 67.18, 68.82)

8 0

3 years ago

F(x)=3x-2<br> h(x)=3x<br> g(x)=x squared<br> find the value of x when f(x)=19

Yuliya22 [10]

The value of f(19) in the given equation is 55.

<h3>What is the function of a linear equation?</h3>

The function of a linear equation illustrates a straight line graph and for a given set of input into the function, there is usually a specific output.

Given that:

f(x) = 3x - 2

We are to find the f(x) = 19.

So;

f(19) = 3(19) - 2

f(19) = 57 - 2

f(19) = 55

Learn more about the function of a linear equation here:

brainly.com/question/14323743

#SPJ1

4 0

2 years ago