De Moivre's theorem uses this general formula z = r(cos α + i<span> sin α) that is where we can have the form a + bi. If the given is raised to a certain number, then the r is raised to the same number while the angles are being multiplied by that number.
For 1) </span>[3cos(27))+isin(27)]^5 we first apply the concept I mentioned above where it becomes
[3^5cos(27*5))+isin(27*5)] and then after simplifying we get, [243 (cos (135) + isin (135))]
it is then further simplified to 243 (-1/ √2) + 243i (1/√2) = -243/√2 + 243/<span>√2 i
and that is the answer.
For 2) </span>[2(cos(40))+isin(40)]^6, we apply the same steps in 1)
[2^6(cos(40*6))+isin(40*6)],
[64(cos(240))+isin(240)] = 64 (-1/2) + 64i (-√3 /2)
And the answer is -32 -32 √3 i
Summary:
1) -243/√2 + 243/√2 i
2)-32 -32 √3 i
The top part equals 1 because anything to the power of zero equals one. then for the second one bring every variable with a negative exponent to the top and leave the 2 at the bottom
Binomial distribution formula: P(x) = (n k) p^k * (1 - p)^n - k
a) Probability that four parts are defective = 0.01374
P(4 defective) = (25 4) (0.04)^4 * (0.96)^21
P(4 defective) = 0.01374
b) Probability that at least one part is defective = 0.6396
Find the probability that 0 parts are defective and subtract that probability from 1.
P(0 defective) = (25 0) (0.04)^0 * (0.96)^25
P(0 defective) = 0.3604
1 - 0.3604 = 0.6396
c) Probability that 25 parts are defective = approximately 0
P(25 defective) = (25 25) (0.04)^25 * (0.96)^0
P(25 defective) = approximately 0
d) Probability that at most 1 part is defective = 0.7358
Find the probability that 0 and 1 parts are defective and add them together.
P(0 defective) = 0.3604 (from above)
P(1 defective) = (25 1) (0.04)^1 * (0.96)^24
P(1 defective) = 0.3754
P(at most 1 defective) = 0.3604 + 0.3754 = 0.7358
e) Mean = 1 | Standard Deviation = 0.9798
mean = n * p
mean = 25 * 0.04 = 1
stdev = 
stdev =
= 0.9798
Hope this helps!! :)
To simplify the function, we need to know some basic identities involving exponents.
1. b^(ax)=(b^x)^a=(b^a)^x
2. b^(x/d) = (b^x)^(1/d) = ((b^(1/d)^x)
Now simplify f(x), where
f(x)=(1/3)*(81)^(3*x/4)
=(1/3)(3^4)^(3*x/4) [ 81=3^4 ]
=(1/3)(3^(4*3*x/4) [ rule 1 above ]
=(1/3) (3^(3*x)
=(1/3)(3^(3x)) [ or (1/3)(27^x), by rule 1 ]
(A) Initial value is the value of the function when x=0, i.e.
initial value
= f(0)
=(1/3)(3^(3x))
=(1/3)(3^(3*0))
=(1/3)(3^0)
=(1/3)(1)
=1/3
(B) the simplified base base is 3 (or 27 if the other form is used)
(C) The domain for an exponential function is all real values ( - ∞ , + ∞ ).
(D) The range of an exponential function with a positive coefficient and without vertical shift is ( 0, + ∞ ).
Answer:
3 days= $105, 6 days= $90, 10 days= $70
Step-by-step explanation:
You can rewrite this equation as y=120-5x
x is the amount of days passed
plug in 3 for x and you get 15. 120-15=105
plug in 6 for x and you get 30. 120-30=90
plug in 10 for x you get 50. 120-50=70
Hope this helps!
:)