Weights measured in grams of randomly selected M&M plain candies:
1 answer:
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.
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Answer: See step by step
Step-by-step explanation: For A my 15 statements are.
- It has 3 triangles inside it, ACD, ADC, and ABC
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- CD≅DB
- D is midpoint of CB
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- Angle BDA equal 90 degrees
- AD≅AD
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-
+
=
12.+
=
13. Triangle ABC has a max of 180 degrees.
14. We can rotate this triangle 180 degrees and it will coincide.
15. We can reflect triangle ACD over vertical line ACD and it will be congruent to ABD.
2. We use pythagorean theorem since it has a right angle.
+
=
Let plug it in.
+
=
1600+ b^2=2025
b^2=424
sqr root of 425 is about 21. Now let find the perimeter.
AB is 45, Since BD+DC=CB, and they are congruent they are equal so 21+21=42 and AC is congruent to AB so it is 45. So the perimeter is 132.
For 3. Start at the orgin, then go up 5 on the y-axis so you should be at (0,5)
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Answer:
The area of the figure is 28 units².
Step-by-step explanation:
Find the area of the full rectangle, ignoring the triangle.
Then, find the area of the triangle:
Finally, subtract: