Answer:
Solution: 9 or more months
Inequality: x > or = 9
Step-by-step explanation:
1. Divide 450 by 52 and you get 8.65...
2. Round up since you can't pay for half a month
Answer:
yes
Step-by-step explanation:
The side lengths satisfy the Pythagorean theorem, so the triangle is a right triangle.
7.5² +10² = 12.5²
56.25 +100 = 156.25
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You may recognize that the ratios of side lengths are ...
7.5 : 10 : 12.5 = 3 : 4 : 5
A 3-4-5 triangle is a well-known right triangle, as this is the smallest set of integers that satisfy the Pythagorean theorem. They also happen to be consecutive integers, so form an arithmetic sequence. Any arithmetic sequence that satisfies the Pythagorean theorem will have these ratios.
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If you're familiar with trigonometry, you know the law of cosines tells you ...
c² = a² + b² - 2ab·cos(θ) . . . . where θ is the angle between sides a and b. This reduces to the Pythagorean theorem when θ=90°, which makes cos(θ)=0. If the sides do not satisfy the Pythagorean theorem, cos(θ)≠0 and the triangle is not a right triangle.
Answer:
least to greatest: {61, 61, 61, 178, 179}
Step-by-step explanation:
If the third-largest angle is 61°, the smallest three angles cannot be larger than 183°. Since the total of all angles must be 540°, and the total of the largest two cannot be greater than 179°×2 = 358°, the sum of the smallest three must be at least 540° -358° = 182°.
So, the possible sets of angles with the smallest 3 totaling 182° or 183° are (in degrees) ...
{60, 61, 61, 179, 179} . . . . two modes
(61, 61, 61, 178, 179} . . . . . one mode -- the set you're looking for
Answer:
15.87% is the chance that Scott takes more than 4.25 minutes to solve a problem at an academic bowl.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4 minutes
Standard Deviation, σ = 0.25 minutes
We standardize the given data.
Formula:
P(more than 4.25 minutes to solve a problem)
Calculation the value from standard normal z table, we have,
Thus,15.87% is the chance that Scott takes more than 4.25 minutes to solve a problem at an academic bowl.
