The number of miles my car can go on a gallon of gasoline is normally distributed with a mean of 30 miles and a standard deviation of 4 miles. i currently have exactly one gallon in my tank, and have to travel 35 miles to my destination.
1. what is the probability i make it to my destination before i run out of gas?
2. find the distance d so that i have a 90% chance of traveling d miles before running out of gas (assuming again that i start with one gallon).
3. suppose i can change my mean mileage per gallon by adjusting the carburetor. what value should i set the mean to so that i have a 90% chance of making it 35 miles on one gallon of gas?

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

b, i believe, but not sure.

the answer would be a 24

step-by-step explanation:

c=2πr=2·π·3=24