I did this test too! The answer was (-9, 0)
I tryed my best to figure out what all of the numbers were. if i got any wrong then please tell me and i will put the right answer in the comments.
1) 3/5= 5/x (cross multiply)
3x=5(5)
3x=25
x= 25/3 -or- 8.33
2) 2/4=x/12<span> (cross multiply</span><span>)
</span> 2(12)=4x
24=4x
6=x
3) 10/x=5/9<span> (cross multiply</span><span>)
10(9)=5x
90=5x
18=x
4) 5/5=x/18</span><span> (cross multiply</span><span>)
5(18)=18x
90=18x
5=x</span>
Answer:
1. 69.4 / 2 = 34.7
2. 38.5 / 2 = 19.25
3. 201.1 / 2 = 100.55
I got these by caculating the area and dividing it by 2
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332
A) The rate for the first 3 years :
15,600 - 7500 = 8100
8100/3 years = 2,700 per year depreciation.
B) The rate between 3 and 8 years:
7500 - 2800 = 4700
4700 / 5 year = 940 per year depreciation.
C) the value of the machine depreciated at a higher rate in the first 3 years. After the first 3 years, the depreciation rate decreased.