The question is incomplete! Complete question along with answers and step by step explanation is provided below.
Question:
(a) Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on each trial. Under what conditions is it appropriate to use a normal approximation to the binomial? (Select all that apply.)
nq > 10
np > 5
p > 0.5
np > 10
p < 0.5
nq > 5
(b) What is the probability of "12" or fewer successes for a binomial experiment with 20 trials. The probability of success on a single trial is 0.50. Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)
Answer:
(a) The correct options are np > 5 and nq > 5
(b) P(x ≤ 12) = 0.8133
Step-by-step explanation:
Please refer to the attached images for explanation, I am unable to type in text editor due to some technical error!
Answer:
33+12t−21t^2
Step-by-step explanation:
(2t-7)²-(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².
4t^2−28t+49−(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².
4t^2−28t+49−(25t^2−40t+16)
To find the opposite of 25t^2
−40t+16, find the opposite of each term.
4t^2−28t+49−25t^2−40t+16
Combine 4t^2 and −25t^2 to get −21t^2.
−21t^2−28t+49+40t−16
Combine −28t and 40t to get 12t.
−21t^2+12t+49−16
Subtract 16 from 49 to get 33.
−21t^2+12t+33
Swap terms to the left side.
33+12t−21t^2
I hope this helped!
Answer:
More informally: The two's complement of an integer is exactly the same thing as its negation. ... It means "to find the negation of a number (i.e., its two's complement) you flip every bit then add 1"
compound interest equation for annually compounded
A=final amount
P=principal
r=rate in decimal
t=time in years
given that
A=1550
P=1000
r=5.5%=0.055
find t
divide both sides by 1000
take ln of both sides
use ln rule 
divide both sides by ln(1.055)
using a calculator, we get that t=8.18544 yrs
so about 8.2yrs
Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = 
= 
= ![2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx](https://tex.z-dn.net/?f=2%5B%20%5Cfrac%7Bx%5E%7B-3%2B1%7D%20%7D%7B-3%2B1%7D%5D%5E%7Bb%7D_a%20%20%20dx)
{ Because ![\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a](https://tex.z-dn.net/?f=%5Cint_%7Ba%7D%5E%7Bb%7D%20x%5E%7Bn%7D%20dx%20%3D%20%5B%20%5Cfrac%7Bx%5E%7Bn%2B1%7D%20%7D%7Bn%2B1%7D%5D%5E%7Bb%7D_a)
}
= ![2[ \frac{x^{-2} }{-2}]^{b}_a](https://tex.z-dn.net/?f=2%5B%20%5Cfrac%7Bx%5E%7B-2%7D%20%7D%7B-2%7D%5D%5E%7Bb%7D_a)
= ![\frac{2}{-2} [ x^{-2} ]^{b}_a](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B-2%7D%20%5B%20x%5E%7B-2%7D%20%5D%5E%7Bb%7D_a)
= ![-1 [ b^{-2} - a^{-2} ]](https://tex.z-dn.net/?f=-1%20%5B%20b%5E%7B-2%7D%20-%20a%5E%7B-2%7D%20%20%5D)
= 
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = 
= 
= 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/
- 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = 
= 
= 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= 
+ (1/
- 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ 
= 0.75
⇒ 
= 1 - 0.75 = 0.25
⇒ 
= 
⇒ = 4 ⇒ x = 
Therefore, value of x such that P(X < x) = 0.75 is 2.
