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.
4 + 3s – 9s = 22
4 - 6s = 22
-6s = 18
-6s/-6 = 18/-6
s = -3
S would equal -3
Best Regards, Mike
<span>1- What is the distance formula?
distance = </span>√(x2-x1)² + (y2-y1)²<span>
2- plug in the correct values from the problem, write the diatance formula with substituted values
</span>distance = √(4-(-3))² + (-6-5)²
<span>
3- simplify the expression, what is the distance between the two points?
distance = </span>√170 = 13.04

They are the same fraction, so they are proportional.
Answer:
C. 132
Step-by-step explanation:
We assume all angle measures are in degrees. The vertical angles are equal in measure, so ...
4x = x +36
3x = 36
x = 12
x +36 = 48
y = 180 -48 = 132
The value of y is 132.