Answer:
The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
Sample mean:

Sample standard deviation:

What is the range of weights of the middle 99.7% of M&M’s?
By the Empirical Rule, within 3 standard deviations of the mean, so:
0.9195 - 3*0.0336 = 0.8187.
0.9195 + 3*0.0336 = 1.0203.
The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.
The answer that appears first is 7
Answer:
A- 2x + 8
Step-by-step explanation:
Combine like terms
6x - 4x = 2x. 2x + 8 is left
Answer:
W = 90 ft
Step-by-step explanation:
1 acre = 43,560 ft^2
A (area) = L (length) x W (width),
L = 484
A = 43560
W = A/L
= 43560/484
= 90 ft
Answer:
28.749 km
Step-by-step explanation:
in reality
1 foot = 0.3048 meters not 0.37
1 mile = 5,280 feet not 5,180
15 miles = 24.14016 kilometers
---------------------------
Using the numbers stated in the problem
0.37m/ft * 5180ft/mile * 15mile * 1km/1000m = 28.749 km