Answer:
I'm pretty sure it's D
Step-by-step explanation:
....
H (-7) = (-7)^2 -5 = 49-5 = 44
Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560
![P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})](https://tex.z-dn.net/?f=P%28X%20%3E%201560%29%20%3D%20P%28%20Z%20%3E%20%5Cdfrac%7Bx%20-%20%5Cmu%7D%7B%5Csigma%7D%29)
![P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})](https://tex.z-dn.net/?f=P%28X%20%3E%201560%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B1560%20-%201450%7D%7B220%7D%29)
![P(X > 1560) = P(Z > \dfrac{110}{220})](https://tex.z-dn.net/?f=P%28X%20%3E%201560%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B110%7D%7B220%7D%29)
P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = ![\sqrt {52 \times 0.3085 (1-0.3085)}](https://tex.z-dn.net/?f=%5Csqrt%20%7B52%20%5Ctimes%200.3085%20%281-0.3085%29%7D)
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)
![Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})](https://tex.z-dn.net/?f=Pr%20%28%20Y%20%3E%2020%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B20.5%20-%2016.042%7D%7B3.3306%7D%29)
![Pr ( Y > 20) = P(Z >1 .338)](https://tex.z-dn.net/?f=Pr%20%28%20Y%20%3E%2020%29%20%3D%20P%28Z%20%3E1%20.338%29)
From z tables
P(Y > 20)
0.0903
Answer:
Assuming each car has the same mass, 694 grams.
Step-by-step explanation:
3470 ÷ 5 = 694