Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is $\hat{p}=26/42=0.6190$ . The estimated standard deviation is given by $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749$ . Because we have a large sample, the 90% confidence interval for p is given by $0.6190\pm z_{0.05}0.0749$ where $z_{0.05}=1.6448$ is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is $0.6190\pm (1.6448)(0.0749)$ , i.e., (0.4958, 0.7422).

5 0

3 years ago

Christine owns 9 hair bands and additional hair bands are 2 for a dollar. How much money does she need to spend on hair bands in

kirill [66]

Answer: $20

Step-by-step explanation:

If she already has 9, and she needs 49, take that extra 9 off 49 and then she needs 40 more and she gets 2 for a dollar, she'll need to spend $20 to get 40 headbands. Can I please get Brainliest?

7 0

3 years ago

C and d please help me im stuck i promise i will give a thank

Troyanec [42]

Answer:

attach the graph in orfer for us to answer

Step-by-step explanation:

3 0

3 years ago

Pls help me with the question?​

Pls help me with the question?