Answer: -5.3k-6
Step-by-step explanation: I dont know if its correct but this is what i got (i hope this helps:)
Answer:
6 races
Step-by-step explanation:
If he won 40% of the races he competed in, and he raced 15 times, this means that he won roughly 4/10 races.
4/10= ?/15
cross multiply 4 and 15 = 60
60/10 = 6
He won 6 races
Give brainliest please!
hope this helps :)
Answer:
a) The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.
b) The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.
Step-by-step explanation:
Given : The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.
To find : What is the probability that
(a) the total gross sales over the next 2 weeks exceeds $5000;
(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?
Solution :
Let
and
denote the sales during week 1 and 2 respectively.
a) Let
Assuming that
and
follows same distribution with same mean and deviation.
![\sigma_X=\sqrt{var(X_1+X_2)}](https://tex.z-dn.net/?f=%5Csigma_X%3D%5Csqrt%7Bvar%28X_1%2BX_2%29%7D)
![\sigma_X=\sqrt{2\sigma^2}](https://tex.z-dn.net/?f=%5Csigma_X%3D%5Csqrt%7B2%5Csigma%5E2%7D)
![\sigma_X=\sqrt{2}\sigma](https://tex.z-dn.net/?f=%5Csigma_X%3D%5Csqrt%7B2%7D%5Csigma)
![\sigma_X=230\sqrt{2}](https://tex.z-dn.net/?f=%5Csigma_X%3D230%5Csqrt%7B2%7D)
So, ![X\sim N(4400,230\sqrt{2})](https://tex.z-dn.net/?f=X%5Csim%20N%284400%2C230%5Csqrt%7B2%7D%29)
![P(X>5000)=1-P(X\leq5000)](https://tex.z-dn.net/?f=P%28X%3E5000%29%3D1-P%28X%5Cleq5000%29)
![P(X>5000)=1-P(Z\leq\frac{5000-4400}{230\sqrt{2}})](https://tex.z-dn.net/?f=P%28X%3E5000%29%3D1-P%28Z%5Cleq%5Cfrac%7B5000-4400%7D%7B230%5Csqrt%7B2%7D%7D%29)
![P(X>5000)=1-P(Z\leq1.844)](https://tex.z-dn.net/?f=P%28X%3E5000%29%3D1-P%28Z%5Cleq1.844%29)
![P(X>5000)=1-0.967](https://tex.z-dn.net/?f=P%28X%3E5000%29%3D1-0.967)
![P(X>5000)=0.0321](https://tex.z-dn.net/?f=P%28X%3E5000%29%3D0.0321)
The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.
b) The probability that sales exceed teh 2000 and amount in at least 2 and 3 next week.
We use binomial distribution with n=3.
![P(X>2000)=1-P(X\leq2000)](https://tex.z-dn.net/?f=P%28X%3E2000%29%3D1-P%28X%5Cleq2000%29)
![P(X>2000)=1-P(Z\leq\frac{2000-2200}{230})](https://tex.z-dn.net/?f=P%28X%3E2000%29%3D1-P%28Z%5Cleq%5Cfrac%7B2000-2200%7D%7B230%7D%29)
![P(X>2000)=1-P(Z\leq-0.87)](https://tex.z-dn.net/?f=P%28X%3E2000%29%3D1-P%28Z%5Cleq-0.87%29)
![P(X>2000)=1-0.1922](https://tex.z-dn.net/?f=P%28X%3E2000%29%3D1-0.1922)
![P(X>2000)=0.808](https://tex.z-dn.net/?f=P%28X%3E2000%29%3D0.808)
Let Y be the number of weeks in which sales exceed 2000.
Now, ![P(Y\geq 2)](https://tex.z-dn.net/?f=P%28Y%5Cgeq%202%29)
So, ![P(Y\geq 2)=P(Y=2)+P(Y=3)](https://tex.z-dn.net/?f=P%28Y%5Cgeq%202%29%3DP%28Y%3D2%29%2BP%28Y%3D3%29)
![P(Y\geq 2)=^3C_2(0.8077)^2\cdot (1-0.8077)+^3C_3(0.8077)^3](https://tex.z-dn.net/?f=P%28Y%5Cgeq%202%29%3D%5E3C_2%280.8077%29%5E2%5Ccdot%20%281-0.8077%29%2B%5E3C_3%280.8077%29%5E3)
![P(Y\geq 2)=0.37635+0.52692](https://tex.z-dn.net/?f=P%28Y%5Cgeq%202%29%3D0.37635%2B0.52692)
![P(Y\geq 2)=0.90327](https://tex.z-dn.net/?f=P%28Y%5Cgeq%202%29%3D0.90327)
The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.
Chris will have to pay back $2,675. you do 2500x7%=175+2500
Answer: f(12) is estimated to 909.09
Step-by-step explanation: