Answer:
The chances Gavis get four or more correct problems is 8/11 or 72.72%
Step-by-step explanation:
The exam is composed of 6 problems out of 12 possible cases (Pc=12). There are 2 groups of problems:
The 8 problems that Gavin has the answer (Problems A).
The 4 problems that Gavin hasn´t the answer (problems B).
Therefore:
P(A≥4)= P(A=4) ∪ P(A=5) ∪ P(A=6) = P(A=4) + P(A=5) + P(A=6)
Before we start analyzing the problem, we have to understand that problems in the exam are selected at random, but a problem can´t be selected twice. therefore picking a specific problem will reduce the pool of that specific group and of the total number of available problems.
If we call
to the probability of an answer of the X group to be the i° picked problem from the j° picked problem of that group:
with
the total number of problems in that group.
We analyze now 3 different problems:
![P(A=6)=A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot A_{55} \cdot A_{66}\\P(A=6)=\frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{3}{7} = \frac{1}{33}](https://tex.z-dn.net/?f=P%28A%3D6%29%3DA_%7B11%7D%20%5Ccdot%20A_%7B22%7D%20%5Ccdot%20A_%7B33%7D%20%5Ccdot%20A_%7B44%7D%20%5Ccdot%20A_%7B55%7D%20%5Ccdot%20A_%7B66%7D%5C%5CP%28A%3D6%29%3D%5Cfrac%7B8%7D%7B12%7D%20%5Cfrac%7B7%7D%7B11%7D%20%5Cfrac%7B6%7D%7B10%7D%20%5Cfrac%7B5%7D%7B9%7D%20%5Cfrac%7B4%7D%7B8%7D%20%5Cfrac%7B3%7D%7B7%7D%20%3D%20%5Cfrac%7B1%7D%7B33%7D)
![P(A=5)=A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot A_{55} \cdot B_{61}+ A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot B_{51} \cdot A_{65}+...+B_{11} \cdot A_{21} \cdot A_{32} \cdot A_{43} \cdot A_{54} \cdot A_{65}\\](https://tex.z-dn.net/?f=P%28A%3D5%29%3DA_%7B11%7D%20%5Ccdot%20A_%7B22%7D%20%5Ccdot%20A_%7B33%7D%20%5Ccdot%20A_%7B44%7D%20%5Ccdot%20A_%7B55%7D%20%5Ccdot%20B_%7B61%7D%2B%20A_%7B11%7D%20%5Ccdot%20A_%7B22%7D%20%5Ccdot%20A_%7B33%7D%20%5Ccdot%20A_%7B44%7D%20%5Ccdot%20B_%7B51%7D%20%5Ccdot%20A_%7B65%7D%2B...%2BB_%7B11%7D%20%5Ccdot%20A_%7B21%7D%20%5Ccdot%20A_%7B32%7D%20%5Ccdot%20A_%7B43%7D%20%5Ccdot%20A_%7B54%7D%20%5Ccdot%20A_%7B65%7D%5C%5C)
![P(A=5)=\frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{4}{7} + \frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{4}{7} + ... + \frac{4}{12} \frac{8}{11} \frac{7}{10} \frac{6}{9} \frac{5}{8} \frac{4}{7} = 6 \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot4 \cdot 4}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}=\frac{8}{33}](https://tex.z-dn.net/?f=P%28A%3D5%29%3D%5Cfrac%7B8%7D%7B12%7D%20%5Cfrac%7B7%7D%7B11%7D%20%5Cfrac%7B6%7D%7B10%7D%20%5Cfrac%7B5%7D%7B9%7D%20%5Cfrac%7B4%7D%7B8%7D%20%5Cfrac%7B4%7D%7B7%7D%20%2B%20%5Cfrac%7B8%7D%7B12%7D%20%5Cfrac%7B7%7D%7B11%7D%20%5Cfrac%7B6%7D%7B10%7D%20%5Cfrac%7B5%7D%7B9%7D%20%5Cfrac%7B4%7D%7B8%7D%20%5Cfrac%7B4%7D%7B7%7D%20%2B%20...%20%2B%20%5Cfrac%7B4%7D%7B12%7D%20%5Cfrac%7B8%7D%7B11%7D%20%5Cfrac%7B7%7D%7B10%7D%20%5Cfrac%7B6%7D%7B9%7D%20%5Cfrac%7B5%7D%7B8%7D%20%5Cfrac%7B4%7D%7B7%7D%20%3D%206%20%5Cfrac%7B8%20%5Ccdot%207%20%5Ccdot%206%20%5Ccdot%205%20%5Ccdot4%20%5Ccdot%204%7D%7B12%20%5Ccdot%2011%20%5Ccdot%2010%20%5Ccdot%209%20%5Ccdot%208%20%5Ccdot%207%7D%3D%5Cfrac%7B8%7D%7B33%7D)
For P(A=4) we can take the solution from P(A=5) and say that:
where "c" is the combinatorial of 2 problems B with 4 problems A. In this case "c" is 15, therefore:
![P(A=4)=\frac{8}{11}](https://tex.z-dn.net/?f=P%28A%3D4%29%3D%5Cfrac%7B8%7D%7B11%7D)
![P(A \ge4)= P(A=4) + P(A=5) + P(A=6) =\frac{1}{33} +\frac{8}{33} +\frac{5}{11} = \frac{8}{11}](https://tex.z-dn.net/?f=P%28A%20%5Cge4%29%3D%20P%28A%3D4%29%20%2B%20P%28A%3D5%29%20%2B%20P%28A%3D6%29%20%3D%5Cfrac%7B1%7D%7B33%7D%20%2B%5Cfrac%7B8%7D%7B33%7D%20%2B%5Cfrac%7B5%7D%7B11%7D%20%3D%20%5Cfrac%7B8%7D%7B11%7D)
-32=2(x^2+10x)
-32=2x^2+20x
-2x^2-20x-32=0
-2(x+2)(x+8)=0
therefore,
x+2=0 or x+8=0
so,...... x= -2 or -8
Answer with explanation:
The given trigonometric Function is,
y= cos x
→The maximum value of , cos x , is 1, at x=0 degree.
Cos x =1
→Cos x = Cos (0)degree,
![x=2 k \pi \pm 0](https://tex.z-dn.net/?f=x%3D2%20k%20%5Cpi%20%5Cpm%200)
→ for, k= An Integer
x=2 k π→→General formula that gives the x-coordinates of the maximum values for y = cos(x)
Ordered Pair =(x,y)=(2 k π, 1)
For, any value of , k in the given equation, y will be always 1.
Answer: The second one
Step-by-step explanation:
Answer:
11042
Step-by-step explanation:
since it says "in all" that means u have to add them up
6,836 + 4,206 = 11,042