Answer:

Step-by-step explanation:
Given



Required
Determine the amount of snow on day 3
This can be calculated using:

Substitute values for the known parameters

Collect Like Terms

Take LCM


Express as mixed number

Answer:
I solved it and my answer is number 1
X-4y=-12
4x-y=12
eliminate y's
multily 2nd equaiton by -4 and add to top one
x-4y=-12
<u>-16x+4y=-48 +
</u>-15x+0y=-60
-15x=-60
divide both sides by -15
x=4
sub back
x-4y=-12
4-4y=-12
minus 4 both sides
-4y=-16
divide both sides by -4
y=4
x=4
y=4
(x,y)
(4,4)<u>
</u>
Answer:
Step-by-step explanation:
Complete Question:
Chapter 6, Section 1-D, Exercise 009 Is a Normal Distribution Appropriate? In each case below, is the sample size large enough so that the sample proportions follow a normal distribution?
a) n=600 p=0.2
b) n=20, p=0.4
if np=10 and npq=10 then the data follows normal distribution
a) np= 120,
q= 1-0.2= 0.8
npq= 600 ×0.2×0.4 = 48
Normal distribution is appropriate and sample size is large enough
b) np= 8
q= 1-0.4= 0.6
npq= 20 × 0.4×0.6= 4.8
sample size is not large enough so normal distribution is not appropriate.
<h3>
Answer:</h3><h3> C</h3><h3 /><h3>
Step-by-step explanation:</h3><h3>It is C.</h3>
Thanks!
:)
Hope i helped