Answer is A and B. Solutions are where the two graphs cross, so you just need to find those coordinates, which are (-5,-13) and (2,-6).
Answer: 8
Step-by-step explanation: Found the ratio of 6/12, which is 1/2, so I used ?/16, plugged it in.
Answer:
its B. -3
Step-by-step explanation:
Remove 3 negative tiles .
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➷ We have 2 points which we can use to figure out the slope
slope = difference in y values/ difference in x values
Substitute in the values you have:
0 - 5/ 3 - 0
slope = -5/3
The y intercept is simply where the line crosses the y axis
In this case, it is 5
Therefore, your equation is y = -5/3x + 5
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a) Add up all the probabilities 
where 
:
b) The expected value is
![E[X]=\displaystyle\sum_xx\,f(x)=3.16](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2Cf%28x%29%3D3.16)
Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.
c) The variance is
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
where
![E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Csum_xx%5E2%5C%2Cf%28x%29%3D11.58)
So the variance is
![V[X]=11.58-3.16^2\approx1.59](https://tex.z-dn.net/?f=V%5BX%5D%3D11.58-3.16%5E2%5Capprox1.59)
The standard deviation is the square root of the variance:
![\sqrt{V[X]}\approx1.26](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%5Capprox1.26)
d) Since 
is a linear combination of 
, computing the expectation and variance of Y is easy:
![E[Y]=E[7X+3]=7E[X]+3=25.12](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5B7X%2B3%5D%3D7E%5BX%5D%2B3%3D25.12)
![V[Y]=V[7X+3]=7^2V[X]\approx78.13](https://tex.z-dn.net/?f=V%5BY%5D%3DV%5B7X%2B3%5D%3D7%5E2V%5BX%5D%5Capprox78.13)
e) The covariance of X and Y is
![\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BX%2CY%5D%3DE%5B%28X-E%5BX%5D%29%28Y-E%5BY%5D%29%5D%3DE%5BXY%5D-E%5BX%5DE%5BY%5D)
We have
so
![E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54](https://tex.z-dn.net/?f=E%5BXY%5D%3DE%5B7X%5E2%2B3X%5D%3D7E%5BX%5E2%5D%2B3E%5BX%5D%3D90.54)
Then the covariance is
![\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BX%2CY%5D%3D90.54-3.16%5Ccdot25.12%5Capprox11.16)
f) Dividing the covariance by the variance of X gives
![\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%7BCov%7D%5BX%2CY%5D%7D%7BV%5BX%5D%7D%5Capprox%5Cdfrac%7B11.16%7D%7B1.59%7D%5Capprox0.9638)
