Probability of success (showing up) = 1-0.05=0.95 is constant and known.
Trials are Bernoulli (show or no show).
Trials are independent and random (assumed from context)
Number of trials is known, n=160.
All being satisfied, we can then model with binomial distribution, where
P(x)=C(n,x)p^x*(1-p)^(n-x)
where C(n,x)=n!/(x!(n-x)!)
Here we look for
P(X<=155)=P(X=0)+P(X=1)+P(X=2)+...+P(X=155)
=0.9061461 (using technology, or add up 156 values calculated, or read from binomial distribution table).
Alternatively, the normal approximation can be used, when n is large.
mean=np=160*0.95=152
standard deviation=sqrt(np(1-p))=2.75681
Apply continuity correction, x=155.5
Z=(155.5-152)/2.75681=1.26958
P(z<=Z)=0.89788 (read from normal distribution tables)
Error=(0.89788-0.9061461)*100%=-0.83%
The approximation is considered good considering p=0.95 is quite skewed, but compensated by n>>50.
Answer:
The original number is 10 times the new number.
Step-by-step explanation:
Answer:
4,5,6 are the three consecutive numbers. 16, 25 and 36 are their squares.
Step-by-step explanation:
Let the three consecutive numbers be x, (x+1), (x+2)
Now, the squares of these three numbers are ![x^{2} ,(x+1)^{2} ,(x+2)^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2C%28x%2B1%29%5E%7B2%7D%20%2C%28x%2B2%29%5E%7B2%7D)
Sum = 77
∴by the problem ,
![x^{2} +(x+1)^{2}+(x+2)^{2} = 77\\x^{2} +(x^{2} +2x+1)+(x^{2}+4x +4) = 77\\x^{2} +x^{2} +2x+1+x^{2} +4x +4 = 77\\3x^{2} +6x+5 = 77\\3x^{2} +6x = 77-5\\3x^{2} + 6x = 72\\3x^{2} +6x-72= 0\\](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%28x%2B1%29%5E%7B2%7D%2B%28x%2B2%29%5E%7B2%7D%20%20%3D%2077%5C%5Cx%5E%7B2%7D%20%2B%28x%5E%7B2%7D%20%2B2x%2B1%29%2B%28x%5E%7B2%7D%2B4x%20%2B4%29%20%3D%2077%5C%5Cx%5E%7B2%7D%20%2Bx%5E%7B2%7D%20%2B2x%2B1%2Bx%5E%7B2%7D%20%2B4x%20%2B4%20%3D%2077%5C%5C3x%5E%7B2%7D%20%2B6x%2B5%20%3D%2077%5C%5C3x%5E%7B2%7D%20%2B6x%20%3D%2077-5%5C%5C3x%5E%7B2%7D%20%2B%206x%20%3D%2072%5C%5C3x%5E%7B2%7D%20%2B6x-72%3D%200%5C%5C)
{Taking 3 common }
![x^{2} +2x- 24 = 0\\](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B2x-%2024%20%3D%200%5C%5C)
{By factorization}
![x^{2} +2x- 24 =0\\ x^{2} +6x-4x-24=0\\x(x+6)-4(x+6)=0\\(x+6)(x-4)=0\\](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B2x-%2024%20%3D0%5C%5C%20x%5E%7B2%7D%20%2B6x-4x-24%3D0%5C%5Cx%28x%2B6%29-4%28x%2B6%29%3D0%5C%5C%28x%2B6%29%28x-4%29%3D0%5C%5C)
Therfore,
![x= -6,4\\](https://tex.z-dn.net/?f=x%3D%20-6%2C4%5C%5C)
<em>X can't be negetive </em>
∴ ![x =4\\x+1=5\\x+2=6](https://tex.z-dn.net/?f=x%20%3D4%5C%5Cx%2B1%3D5%5C%5Cx%2B2%3D6)
The squares of the three consecutive numbers are 16, 25, 36
The three consecutive numbers whose sum is 77 are 4, 5, 6
Answer:
1/6
Step-by-step explanation:
1/3 + /1/6= 1/2
Because 1.5 is half of 3
Answer: 11+5=16
Step-by-step explanation: 14, 15, 16, 17 and 18