Answer:
In the case of a Type I error, the null hypothesis would be wrongly rejected and the school district will conclude that the new technology is effective when it is not.
They will start to pay for the software when in fact it does not improve Algebra 1 skills.
Step-by-step explanation:
A Type I error happens when a true null hypothesis is rejected.
The probability of a Type I error is equal to the significance level, as it is the probabilty of getting an sample result with low probability but only due to chance, as the null hypothesis is in fact true.
In this scenario, the null hypothesis would represent the claim that the new technology does not make significant improvement.
In the case of a Type I error, this null hypothesis would be wrongly rejected and the school district will conclude that the new technology is effective when it is not.
They will start to pay for the software when in fact it does not improve Algebra 1 skills.
With a mean of λ , the probability mass distribution (pmf) is given by
![P(k,\lambda)=\lambda^k*e^(-\lambda)/k!](https://tex.z-dn.net/?f=P%28k%2C%5Clambda%29%3D%5Clambda%5Ek%2Ae%5E%28-%5Clambda%29%2Fk%21)
for λ = 4, and k<2 (i.e. k=0 or 1)
P(k<2, λ )=P(k=1, λ ) + P(k=1, λ )
![=4^0*e^(-4)/4!+4^1*e^(-4)/1!](https://tex.z-dn.net/?f=%3D4%5E0%2Ae%5E%28-4%29%2F4%21%2B4%5E1%2Ae%5E%28-4%29%2F1%21)
![=4^0*e^(-4)/4!+4^1*e^(-4)/1!](https://tex.z-dn.net/?f=%3D4%5E0%2Ae%5E%28-4%29%2F4%21%2B4%5E1%2Ae%5E%28-4%29%2F1%21)
=0.01832+0.07326
=0.09158 (to the fifth place of decimal)
Note: Poisson processes have no memory, so 2 calls in first hour has the same probability as 2 calls in any other hour.
X*0.2=y
if x = the total commissions, and 20% goes to his assistant, you would need to multiply your total sales commissions by 0.2 to find 20%, and that would be the total given to the assistant.
Hope this helps
3a^6 -2a^5 +a^4
Brainly says my answer is too short so I have to add extra text here
Answer:
I totally agree
Step-by-step explanation:
It helps a lot.