Answer:
90%
Step-by-step explanation:
Let R = 100
Since, Z = 30% of R
Therefore, Z = 30% *100 = 30
Since, W = 3Z
Therefore, W = 3*30 = 90
Percentage of W of R

9+9+9+9+9=45 I'm not sure if that's what they're asking for
Answer:
Yes, it would be statistically significant
Step-by-step explanation:
The information given are;
The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%
Number of jawbreakers in the sample, n = 800
The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 =
=p
The formula for the standard deviation of a proportion is 
Solving for the standard deviation gives;

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4 in the sample of 800 = 800*0.6 = 480
For statistical significance the difference from the mean = 2×
= 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7
The z-score of 494 jawbreakers is given as follows;


Therefore, the z-score more than 2 ×
which is significant.