Answer:
Shortest Leg: 5 inches, Hypotenuse: 13 inches
Step-by-step explanation:
A right triangle follows the pythagorean theorem, a^2+b^2=c^2. C is the hypotenus. A combination found using this is 5, 12, and 13. To double check this, we know that the hypotenus is 2x+3, where x is the smallest leg.
32 degees, 43.8 degrees, 53.8, 63.8 Yes they are proportional
- Midpoint Formula:
So firstly, let's start with the x-coordinates. Since we know the midpoint's x-coordinate and point A's x-coordinate, we can solve for point B's x-coordinate as such:
Next, do the same thing except solve for the y-coordinate and using point A's y-coordinate and the midpoint's y-coordinate:
<u>Putting it together, point B's coordinates are (2,4).</u>
The volume of soup in the cylindrical can is 100.48 inches cube.
<h3>How to find the volume of a cylindrical can?</h3>
We have to find the volume of the soup in a cylindrical can of height 8 inches and 4 inches across the lid.
The volume of the soup is the volume of the cylindrical can.
Therefore,
volume of the cylindrical can = πr²h
where
- r = radius of the cylinder
- h = height of the cylinder
Therefore,
h = 8 inches
r = 4 / 2 = 2 inches
volume of the soup in the cylindrical can = πr²h
volume of the soup in the cylindrical can = π × 2² × 8
volume of the soup in the cylindrical can = π × 4 × 8
volume of the soup in the cylindrical can = 32π
Therefore,
volume of the soup in the cylindrical can = 32 × 3.14
volume of the soup in the cylindrical can = 100.48 inches³
learn more on volume here:brainly.com/question/23207959
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Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
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My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
