Answer:No
Step-by-step explanation:
Given system of equation is:
2x + 3 = y
2x + y = 15
To check whether (2,7) is solution to this system or not, we will put x=2 and y=7 in both equations.
Putting x=2 and y=7 in Eqn 1
2(2) + 3 = 7
4 + 3 = 7
7 = 7
Thus the ordered pair satisfies the equation
Putting x=2 and y=7 in Eqn 2
2(2) + 7 = 15
4 + 7 = 15
11 ≠ 15
The ordered pair do not satisfy the second equation.
Hence,
(2,7) is not a solution to the given system of equations.
Answer:
-7/4
Step-by-step explanation:
You are looking for the composite g(f(2)). The simplest way to solve this is to evaluate f(2) and enter the solution in to your g function.
g(f(2))=g(-(2)^2-2(2)+4)=g(-4-4+4)=g(-4)
g(-4)=4/(-4(-4)-2)=4/(16-2)=4/14=2/7
Therfor, g(f(2))=2/7 **I'm assuming the -4x-2 is all in the denominator of the g(x) function. If -2 is not in the denominator you would have
g(f(2))=4/(-4(-4)) -2=4/16 -2=1/4 -2=1/4-8/4= -7/4
Matt could do one of the following arrangements;
2 rows of 8 CDs
4 rows of 4 CDs
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
