Answer:
The answer to your question is (3, 0)
Step-by-step explanation:
Data
4x + 2y = 12 Equation l
x - y = 3 Equation ll
Process
To solve this problem use the elimination method
1.- Multiply equation ll by 2
4x + 2y = 12
2x - 2y = 6
6x 0 = 18
2.- Solve for x
6x = 18
x = 18/6
x = 3
3.- Substitute x in equation ll to find y
3 - y = 3
-Solve for y
-y = 3 - 3
y = 0
4.- Write the solution
(3, 0)
I think x = 2 is the answer
Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
![\left[\begin{array}{cc}Savings&Frequency\\\$0-\$199&339\\\$200-\$399&86\\\$400-\$599&55\\\$600-\$799&18\\\$800-\$999&11\\\$1000-\$1199&8\\\$1200-\$1399&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DSavings%26Frequency%5C%5C%5C%240-%5C%24199%26339%5C%5C%5C%24200-%5C%24399%2686%5C%5C%5C%24400-%5C%24599%2655%5C%5C%5C%24600-%5C%24799%2618%5C%5C%5C%24800-%5C%24999%2611%5C%5C%5C%241000-%5C%241199%268%5C%5C%5C%241200-%5C%241399%263%5Cend%7Barray%7D%5Cright%5D)
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
<span>The computation for the confidence level = (18/20) x 100% =
90 %, the E or the margin error = 0.023 and the p = to 0.71. Confidence
interval can be achieved by using the formula (p-E, p+E) = (0.71-0.025,
0.71+0.025). Therefore, the confidence interval is (0.685,0.735).</span>