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.
Yo sup??
we can solve this question by applying trigonometric ratios
cos59=CB/CD
CD=CB/cos59
=7.76
=7.8
Hope this helps.
Not sure if you mean “Round 37 to the nearest tenth” but if so then it’s “40”.
(-7-6) - 16 = −29
(4+6) - 15 = −5
14/3 + 2/9 = 44/9
8/3 x 1/5 = 8/15
3/4 % 8/15 = 1/250
<em>please</em><em> mark me </em><em>brainliest</em><em> if someone else answers!</em>
The answer should be the third one I believe