The number of blues was two more than the total number of reds and greens. Three times the number of reds added to twice the number of greens was twice the number of blues.
If there were 22 total, how many were there of each color?

Solved

Show answers

Ответ:

aubford9442

03.07.2022

Mathematics

98% Confidence interval: (0.5483,0.6891)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 257

Number of adults total body length was 28 feet or longer, x = 159

$\hat{p} = \dfrac{x}{n} = \dfrac{159}{257} = 0.6187$

98% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.02} = 2.326$

Putting the values, we get:

$0.6187\pm 2.326(\sqrt{\dfrac{0.6187(1-0.6187)}{257}})\\\\ = 0.6187\pm 0.0704\\\\=(0.5483,0.6891)$

is the required 98% confidence interval.

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

I want advice

The number of blues was two more than the total number of reds and greens. Three times the number of reds added to twice the number of greens was twice the number of blues. If there were 22 total, how many were there of each color?