Explanation:
First we need to find the amount of the diet that is fruit. (It is what is not biscuits or bamboo.) The quantity is given per day, so we need to multiply that by the number of days in 3 weeks.
(1 - 23% -73%)·(1100 g/day)·(7 day/week)·(3 week) = (44 g/day)·(21 day)
= 924 g . . . . of fruit
Answer: 90% confidence interval is; ( - 0.0516, 0.3752 )
Step-by-step explanation:
Given the data in the question;
n1 = 72, n2 = 17
P1 = 54 / 72 = 0.75
P2 = 10 / 17 = 0.5882
so
P_good = 0.75
P_bad = 0.5882
standard ERRROR will be;
SE = √[(0.75×(1-0.75)/72) + (0.5882×(1-0.5882)/17)]
SE = √( 0.002604 + 0.01424)
SE = 0.12978
given confidence interval = 90%
significance level a = (1 - 90/100) = 0.1, |Z( 0.1/2=0.05)| = 1.645 { from standard normal table}
so
93% CI is;
(0.75 - 0.5882) - 1.645×0.12978 <P_good - P_bad< (0.75 - 0.5882) + 1.645×0.12978
⇒0.1618 - 0.2134 <P_good - P_bad< 0.1618 + 0.2134
⇒ - 0.0516 <P_good - P_bad< 0.3752
Therefore 90% confidence interval is; ( - 0.0516, 0.3752 )
Answer:
parallel lines
Step-by-step explanation:
the slope is 0
Answer:
The answer is 62 = 2 × 31...........