$m\widehat {BCD} = 2\times m\angle BAD\\\therefore m\widehat {BCD} = 2\times 129\degree \\\therefore m\widehat {BCD} = 258\degree\\\\\because m\widehat{CD} = m\widehat {BCD}-m\widehat{BC} \\\therefore m\widehat{CD} = 258\degree - 146\degree \\\huge \red {\boxed {\therefore m\widehat{CD} = 112\degree}}$

6 0

3 years ago

A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 80% confidence inter

jasenka [17]

Answer:

The 80% confidence interval for the mean

(4.6199 , 4.7801)

Step-by-step explanation:

Explanation:-

Assuming that the population standard deviation for the number of energy drinks consumed each week is 1

Given the Population standard deviation 'σ' = 1

The study found that for a sample of 256 teenagers the mean number of energy drinks consumed per week is 4.7

given sample size 'n' = 256

mean of the sample 'x⁻' = 4.7

confidence interval for the mean

The 80% confidence interval for the mean is determined by

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} + Z_{\alpha }\frac{S.D}{\sqrt{n} } )$

the z-score of 80% level of significance = 1.282

$(4.7 - 1.282\frac{1}{\sqrt{256} } , 4.7 + 1.282\frac{1}{\sqrt{256} } )$

(4.7 - 0.0801 , 4.7 +0.0801)

(4.6199 , 4.7801)

Conclusion:-

The 80% confidence interval for the mean

(4.6199 , 4.7801)

7 0

4 years ago