In order for the mode to be 6, there must be more of them than any other number. In order for the mean to be 3.5, most of the distribution must lie at 3 or below to balance the large number of 6s. For the median to be 3, there must be as many scores at or above 3 as at or below 3. Thus, we choose to have most of the distribution with values 1, 2, 3 and only enough 6s to make that be the mode.
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 )
43 tens = 43 * 10 = 430
3 tens = 3 * 10 = 30
430 x 30 = 12900 tens
12900/100 = 129 hundreds <===
Answer:
7 i think not sure
Step-by-step explanation:
Step-by-step explanation:
always write what you know
circumference
U = 2 x pi x r
r = d/2
U = 39.25
rearrange and solve for r
U / (2 x pi) = r
solve for diameter
2r = d
