Answer:
800%
Step-by-step explanation:
<h2><u>Percentage increase</u></h2><h3><u>formula:</u></h3>
change/ original * 100
change is 1.35 - 0.15 = 1.2
original is the amount in the beginning which is 0.15
= 1.2/0.15 * 100
=800 %
<em><u>the percentage increase is by 800%</u></em>
There are two ways to work this out: normal variables or using "imaginary" numbers.
Normal variables:
![(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}](https://tex.z-dn.net/?f=%20%287%2B2i%29%283-i%29%5C%5C%287%2A3%29%2B%5B7%2A%28-i%29%5D%2B%283%2A2i%29%2B%5B2i%2A%28-i%29%5D%5C%5C21-7i%2B6i-2i%5E%7B2%7D%5C%5C%5C%5C21-i-2i%5E%7B2%7D)
Imaginary numbers:
Using the result from earlier:
Now since
, then the expression becomes:
It is y = 3x + 2, which means the y-intercept is 2. :)
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
Answer:
5
Step-by-step explanation:
Let's call the length AD y.
Hope this helps!
