Answer: 2.4
Step-by-step explanation:
Given : The weights, in pounds, of the cats in an animal adoption center are normally distributed.
The confidence interval is (7.9,12.7) (1)
Let be the sample mean of the weights, in pounds, of the cats in an animal adoption center.
We know that the confidence interval for population mean is given by :-
(2)
From (1) and (2)
Subtracting (3) from (4), we get
Hence, the margin of error is E = 2.4
Answer:
Check pdf
Step-by-step explanation:
Answer:
d) The additional training did not significantly lower the defect rate
Step-by-step explanation:
Let proportion of defective chips be = x
Null Hypothesis [H0] : Additional training has no impact on defect rate x = 8% = 0.08
Alternate Hypothesis [H1] : Additional training has impact on defect rate x < 8% , x < 0.08
Observed x proportion (mean) : x' = 27 / 450 = 0.06
z statistic = [ x' - x ] / √ [ { x ( 1-x ) } / n ]
( 0.06 - 0.08 ) / √ [ 0.08 (0.92) / 450 ]
= -0.02 / √ 0.0001635
= -0.02 / 0.01278
z = - 1.56
Since calculated value of z, 1.56 < tabulated value of z at assumed 0.01 significance level, 2.33
Null Hypothesis is accepted, 'training didn't have defect rate reduction impact' is concluded
Answer: the answer is 19
Step-by-step explanation:okay so she already have read 119 pages and the total pages is 271 so we substract 271-119=152 so she has 152 pages left so we divide 152 pages divided by 8 because she read 8 pages per hour so 152/8=19
Answer:
It's 21, you are correct!!
Step-by-step explanation:
Answer:
26 whole 3/40
Step-by-step explanation:
Multiply the denominator with the number in middle and add to top,
13/3+ 17/5+ 46/16
LCM& Changing into lowest terms,
(65+51)/15+23/8
116/5+23/8
(928+115)/40
1043/40
26 whole 3/40
Answer:
167
Step-by-step explanation: