Answer: 0.0227502
Step-by-step explanation:
Let x denote the random variable that represents the mileage of SUV.
As per given we have,


sample size : n= 36
We assume that the mileage of SUV is normally distributed.
Z-score value corresponds to x= 21.0,

Using standard z-value table ,
The probability that the sample average will be over 21.0 mpg:-
P(X>35)=P(z>2)=1-P(z<2)=1-0.9772498=0.0227502
Hence, the required probability = 0.0227502
Answer:
Step-by-step explanation:
Te processing fee was applied to the discounted price.
The discounted price is
- 78 - 40% =
- 78*(100 - 40)/100 =
- 78*0.6 =
- 46.8
Amount of fee is
- 20% of 46.8 =
- 0.2*46.8 =
- 9.36
2= 2*1
6= 2*3
8= 2*4
8= 2*2*2
The lcm of these three numbers equals 2*2*2*3*1. Notice how we only multiply 3 sets of 2. This is because 2 is a common factor between all 3 of the numbers.
2*2*2*3*1= 24
Final answer: 24
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