Suppose a runner completes one lap around a 400-m track in a time of 50 s. calculate the magnitude of the average velocity of th
2 answers:
Step-by-step explanation:
It is given that,
Distance covered by the runner, d = 400 m
Time taken to complete one lap, t = 50 s
We need to find the magnitude of the average velocity of the runner. Total displacement divided by total time taken is called the average velocity of the runner. Here, the total displacement is 0 as runner come back to same point.
It is given by :
v = 0
So, the average velocity of the runner is 0. Hence, this is the required solution.
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Answer:
- 8
Step-by-step explanation:
Given
(3x² - 5)(4 + 4x²)
Each term in the second factor is multiplied by each term in the first factor, that is
3x²(4 + 4x²) - 5 (4 + 4x²) ← distribute both parenthesis
= 12x² + 12 - 20 - 20x² ← collect like terms
= 12 - 8x² - 20
The coefficient of the x² term is - 8
The value of sin(30°) is: One-half ⇒ 2nd answer
Step-by-step explanation:
In a right triangle there are two acute angles, the side opposite to the right angle is called hypotenuse, and the other two sides are opposite and adjacent to the acute angles
- sine the acute angle (sin) = opposite side to it/hypotenuse
- cosine the acute angle (cos) = adjacent side to it/hypotenuse
- Tangent the acute angle (tan) = opposite side to it/adjacent side to it
In Δ QRS:
∵ m∠QRS = 90°
∵ SQ is opposite to ∠QRS
∴ SQ is the hypotenuse
∵ SQ = 10 units
∴ The hypotenuse = 10
∵ m∠RSQ = 30°
- The opposite side to ∠RSQ is RQ
∵ RQ = 5 units
∴ The opposite side to the angle of 30° = 5
∵ sin(30°) = opposite side to 30°/hypotenuse
∵ The opposite side to angle 30° = 5
∵ The hypotenuse = 10
∴ sin(30°) =
∴ sin(30°) =
The value of sin(30°) is: One-half
Learn more:
You can learn more about the trigonometry ratios in brainly.com/question/9880052
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Answer:
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549
Step-by-step explanation:
We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.0930
The margin of error is:
M = T*s = 2.0930*0.16 = 0.3349.
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.92 - 0.3349 = 2.5851
The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549
