<span>n = 5
The formula for the confidence interval (CI) is
CI = m ± z*d/sqrt(n)
where
CI = confidence interval
m = mean
z = z value in standard normal table for desired confidence
n = number of samples
Since we want a 95% confidence interval, we need to divide that in half to get
95/2 = 47.5
Looking up 0.475 in a standard normal table gives us a z value of 1.96
Since we want the margin of error to be ± 0.0001, we want the expression ± z*d/sqrt(n) to also be ± 0.0001. And to simplify things, we can omit the ± and use the formula
0.0001 = z*d/sqrt(n)
Substitute the value z that we looked up, and get
0.0001 = 1.96*d/sqrt(n)
Substitute the standard deviation that we were given and
0.0001 = 1.96*0.001/sqrt(n)
0.0001 = 0.00196/sqrt(n)
Solve for n
0.0001*sqrt(n) = 0.00196
sqrt(n) = 19.6
n = 4.427188724
Since you can't have a fractional value for n, then n should be at least 5 for a 95% confidence interval that the measured mean is within 0.0001 grams of the correct mass.</span>
First, convert 3 into a fraction with denominator 8.
(3 pizzas)(8/8 pizzas)=24/8 pizzas
Then, divide 24/8 pizzas by 2/8 pizzas per 1 person.
(24/8 pizzas)(2/8 pizzas/person)
Dividing by a fraction is the same as multiplying by the reciprocal:
(24/8)/(2/8)=(24/8)×(8/2) the 8s cancel and you are left with 24/2=12
3 pizzas will feed 12 people.
SO YOU DO 72 MULTIPLIED BY 5 AND YOU WILL GET YOUR ANSWER
I did out all my work but it got deleted. Hope you don't mind just answers.
5) 10
6) 32
7) 64 pi
8) A=64 pi C=16 pi
9) A=110.25 pi C=21 pi
10) A=1024/pi C=64
11) L=4sqrt(5) or about 9 midpoint is (1,4)
12) L=sqrt(37) or about 6 midpoint is (-1,-3.5)
Sorry I couldn't show work, hope this helps :)
