Answer:
The number is -1/2.
Step-by-step explanation:
I am assuming they are fractions. So we have the equation:
5/2 x + 2/3 = -7/12
We multiply through by 12 to eliminate the fractions:
5/2 * 12 x + 2/3 * 12 = -7/12 * 12
30x + 8 = -7
30x = -15
x = -15/30
x = -1/2.
Sample space ={5G, 2Y, 8R, 3P} =18 POSSIBLE OUTCOME
1st draw: P(R) = 8/18 = 4/9 = 0.444
2nd draw (no replacement, means already one Red is picked up, so the sample space has been reduced): P(another one more RED) = 7/17 =0.411
Answer:
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Step-by-step explanation:
Sample size, n = 51
p = 0.62
1 - p = 1 - 0.62 = 0.38
n = 515
Confidence level = 90% = Zcritical at 90% = 1.645
Confidence interval = (p ± margin of error)
Margin of Error = Zcritical * sqrt[(p(1-p))/n]
Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]
Margin of Error = 1.645 * 0.0214
Margin of Error = 0.035203
Lower boundary = (0.62 - 0.035203) = 0.584797
Upper boundary = (0.62 + 0.035203) = 0.655203
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
-6-7(c+10)
-6-7c-70
-76-7c
B
