Following are the solution parts for the given question:
For question A:
In the given question, we calculate 
of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:
Using the t table we calculate 
When of the confidence interval:
So confidence interval for the mean weight of shipped homemade candies is between 
.
For question B:
Here we need to calculate confidence interval for the true proportion of all college students who own a car which can be calculated as
Using the Z-table we found 
therefore the confidence interval for the genuine proportion of college students who possess a car is
So the confidence interval for the genuine proportion of college students who possess a car is between 
For question C:
- In question A, We are certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
-
In question B, We are positive that the true percentage of college students who possess a car is between 0.28 and 0.34.
Learn more about confidence intervals:
brainly.in/question/16329412
A. you plug in to a calculator, which will give 1840.986 so you need to round up to 1840.99. if you truncate it to .98 then he won't reach 2000 in 3 years
b. for this one if you look at the equation given to find the principle it is principle = result (1+rate) ^ -time
if you re arrange this you get result=principle (1+rate)^time
so result = 1840.99(1.028)^5
= 2113.57
Answer:
ummm what is this for tho?
Answer:
Write in slope-intercept form,
y =mx+bxy=4x+6
Step-by-step explanation:
-24a²- 6a +9 = - 3(8a² +6a - 9)
8a² +6a - 9 =0
x=(-b +/-√(b² - 4ac))/2a
x = (-6 +/-√(36+4*8*9) /(2*8) = (-6 +/-√(324) /16 = (-6 +/-18)/16
x1 = (-6 +18)/16 = 0.75
x2 = (-6 -18)/16 = - 1.5
-24a²- 6a +9 = - 3(8a² +6a - 9) = -3(a - 0.75)(a + 1.5)
