Jerry’s cost is $.04 per message
Answer:
C. 5 weeks.
Step-by-step explanation:
In this question we have a random variable that is equal to the sum of two normal-distributed random variables.
If we have two random variables X and Y, both normally distributed, the sum will have this properties:
![S=X+Y\\\\\ \mu_S=\mu_X+\mu_Y=30+60=90\\\\\sigma_S=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}=22.36](https://tex.z-dn.net/?f=S%3DX%2BY%5C%5C%5C%5C%5C%20%5Cmu_S%3D%5Cmu_X%2B%5Cmu_Y%3D30%2B60%3D90%5C%5C%5C%5C%5Csigma_S%3D%5Csqrt%7B%5Csigma_X%5E2%2B%5Csigma_Y%5E2%7D%3D%5Csqrt%7B10%5E2%2B20%5E2%7D%3D%5Csqrt%7B100%2B400%7D%3D%5Csqrt%7B500%7D%3D22.36)
To calculate the expected weeks that the donation exceeds $120, first we can calculate the probability of S>120:
![z=\frac{S-\mu_S}{\sigma_S} =\frac{120-90}{22.36}=\frac{30}{22.36}=1.34\\\\P(S>120)=P(z>1.34)=0.09012](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BS-%5Cmu_S%7D%7B%5Csigma_S%7D%20%3D%5Cfrac%7B120-90%7D%7B22.36%7D%3D%5Cfrac%7B30%7D%7B22.36%7D%3D1.34%5C%5C%5C%5CP%28S%3E120%29%3DP%28z%3E1.34%29%3D0.09012)
The expected weeks can be calculated as the product of the number of weeks in the year (52) and this probability:
![E=\#weeks*P(S>120)=52*0.09012=4.68](https://tex.z-dn.net/?f=E%3D%5C%23weeks%2AP%28S%3E120%29%3D52%2A0.09012%3D4.68)
The nearest answer is C. 5 weeks.
Answer:
Hey Friend..... can't see the attachment....
Mr. Valentino can`t go over his budget of $25. The amount of money times the amount of boxes needs to be less than or equal to $25:
Answer: A ) $1.75 x ≤ $25,
where x is the amount of boxes he bought.
Answer:
3156
Step-by-step explanation:
- <em>Used formula:</em>
- <em>(1² + 2² + 3² + ... + n²) =1/6*n(n + 1)(2n + 1)</em>
--------
- 10²+12²+14²+......+26² =
- (2*5)²+(2*6)² + (2*7)² + ... + (2*13)² =
- 4*(5²+6²+7²+...+13²) =
- 4*(1²+2²+...+13² - (1²+2²+3²+4²)) =
- 4*(1/6*13(13+1)(2*13+1) - (1+4+9+16)) =
- 4*(1/6*13*14*27- 30) =
- 4*(819 - 30) =
- 4*789 =
- 3156