No! 11, the input, has two outputs which makes it not a function! :)
Yes, i have to make this answer longer so this is- you know.
The total number of pencils is 2,160. When estimating to the nearest hundredth one gets 2,200, which is an underestimate.
The advantage of underestimating is that your answer is more accurate than an overestimate.
Answer:77
Step-by-step explanation:
Least Common Multiple of 7 and 11 with GCF Formula
The formula of LCM is LCM(a,b) = ( a × b) / GCF(a,b).
We need to calculate greatest common factor 7 and 11, than apply into the LCM equation.
GCF(7,11) = 1
LCM(7,11) = ( 7 × 11) / 1
LCM(7,11) = 77 / 1
LCM(7,11) = 77
Least Common Multiple (LCM) of 7 and 11 with Primes
Least common multiple can be found by multiplying the highest exponent prime factors of 7 and 11. First we will calculate the prime factors of 7 and 11.
Prime Factorization of 7
Prime factors of 7 are 7. Prime factorization of 7 in exponential form is:
7 = 71
Prime Factorization of 11
Prime factors of 11 are 11. Prime factorization of 11 in exponential form is:
11 = 111
Now multiplying the highest exponent prime factors to calculate the LCM of 7 and 11.
LCM(7,11) = 71 × 111
LCM(7,11) = 77
Answer:
P(success)=0.2 matches the probability histogram.
Step-by-step explanation:
Formula : ![P(X)=^nC_r p^r(1-p)^{n-r}](https://tex.z-dn.net/?f=P%28X%29%3D%5EnC_r%20p%5Er%281-p%29%5E%7Bn-r%7D)
Where X represents the number of successes.
p represents probability of success .
1-p represents probability of failure.
Now when X= 0 the probability is 0.64 (Using histogram )
So, ![^2C_0 p^0(1-p)^{2}=0.64](https://tex.z-dn.net/?f=%5E2C_0%20p%5E0%281-p%29%5E%7B2%7D%3D0.64)
![(1-p)^{2}=0.64](https://tex.z-dn.net/?f=%281-p%29%5E%7B2%7D%3D0.64)
![(1-p)=\sqrt{0.64}](https://tex.z-dn.net/?f=%281-p%29%3D%5Csqrt%7B0.64%7D)
![(1-p)=0.8](https://tex.z-dn.net/?f=%281-p%29%3D0.8)
![1-0.8=p](https://tex.z-dn.net/?f=1-0.8%3Dp)
![0.2=p](https://tex.z-dn.net/?f=0.2%3Dp)
So, the probability of success = 0.2
Hence P(success)=0.2 matches the probability histogram.